Differential Equations — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 9, Differential Equations (অৱকল সমীকৰণ). It covers Exercise 9.1, Exercise 9.2, Exercise 9.3, Exercise 9.4, Exercise 9.5 and the Miscellaneous Exercise on Chapter 9, with each answer fully worked out so you can follow every step and prepare confidently for your examination.
Summary
A differential equation is an equation that involves the derivatives of a dependent variable with respect to one or more independent variables. The order of a differential equation is the order of the highest-order derivative appearing in it, and its degree (defined only when the equation is a polynomial in the derivatives) is the highest power of that highest-order derivative. For example, $\frac{d^2y}{dx^2} + y = 0$ has order $2$ and degree $1$, while $y^{\prime\prime\prime} + \sin(y^{\prime\prime}) = 0$ has order $3$ but no defined degree.
A function that satisfies a differential equation is called its solution. A solution containing as many arbitrary constants as the order of the equation is a general solution, and one free from arbitrary constants (obtained by giving the constants particular values) is a particular solution. The general solution therefore represents a whole family of curves, and a particular solution is a single member of that family, as illustrated below.
Three methods are used to solve first-order, first-degree equations. In the variable-separable method the equation is written as $\frac{1}{h(y)}\,dy = g(x)\,dx$ and both sides are integrated. A homogeneous equation $\frac{dy}{dx} = F(x, y)$, where $F$ is homogeneous of degree zero, is solved by the substitution $y = vx$. A linear equation $\frac{dy}{dx} + Py = Q$ is solved using the integrating factor $\text{I.F.} = e^{\int P\,dx}$, giving $y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.})\,dx + C$.
Summary: ASSEB Class 12 Mathematics Chapter 9 Differential Equations explains the order and degree of a differential equation, general and particular solutions, and the three standard methods — variables separable, homogeneous equations solved by $y = vx$, and linear equations solved with the integrating factor $e^{\int P\,dx}$ — with complete worked answers to Exercise 9.1, 9.2, 9.3, 9.4, 9.5 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 9.1
Determine order and degree (if defined) of the differential equations given in Exercises 1 to 10.
1. $\dfrac{d^4y}{dx^4} + \sin(y^{\prime\prime\prime}) = 0$
Answer: The highest-order derivative present is $\dfrac{d^4y}{dx^4}$, so the order is $4$. Because the term $\sin(y^{\prime\prime\prime})$ makes the equation non-polynomial in its derivatives, the degree is not defined.
2. $y^{\prime} + 5y = 0$
Answer: The highest-order derivative is $y^{\prime} = \dfrac{dy}{dx}$, so the order is $1$. It is a polynomial in $y^{\prime}$ and the highest power of $y^{\prime}$ is $1$, so the degree is $1$.
3. $\left(\dfrac{ds}{dt}\right)^4 + 3s\,\dfrac{d^2s}{dt^2} = 0$
Answer: The highest-order derivative is $\dfrac{d^2s}{dt^2}$, so the order is $2$. The equation is a polynomial in the derivatives and the highest power of $\dfrac{d^2s}{dt^2}$ is $1$, so the degree is $1$.
4. $\left(\dfrac{d^2y}{dx^2}\right)^2 + \cos\left(\dfrac{dy}{dx}\right) = 0$
Answer: The highest-order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. Because of the term $\cos\left(\dfrac{dy}{dx}\right)$ the equation is not a polynomial in its derivatives, so the degree is not defined.
5. $\dfrac{d^2y}{dx^2} = \cos 3x + \sin 3x$
Answer: The highest-order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. Writing it as $\dfrac{d^2y}{dx^2} – \cos 3x – \sin 3x = 0$, it is a polynomial in the derivative with the highest power of $\dfrac{d^2y}{dx^2}$ equal to $1$, so the degree is $1$.
6. $(y^{\prime\prime\prime})^2 + (y^{\prime\prime})^3 + (y^{\prime})^4 + y^5 = 0$
Answer: The highest-order derivative is $y^{\prime\prime\prime}$, so the order is $3$. It is a polynomial in the derivatives and the highest power of $y^{\prime\prime\prime}$ is $2$, so the degree is $2$.
7. $y^{\prime\prime\prime} + 2y^{\prime\prime} + y^{\prime} = 0$
Answer: The highest-order derivative is $y^{\prime\prime\prime}$, so the order is $3$. It is a polynomial in the derivatives with the highest power of $y^{\prime\prime\prime}$ equal to $1$, so the degree is $1$.
8. $y^{\prime} + y = e^x$
Answer: The highest-order derivative is $y^{\prime}$, so the order is $1$. It is a polynomial in $y^{\prime}$ with the highest power $1$, so the degree is $1$.
9. $y^{\prime\prime} + (y^{\prime})^2 + 2y = 0$
Answer: The highest-order derivative is $y^{\prime\prime}$, so the order is $2$. It is a polynomial in the derivatives; the highest power of the highest-order derivative $y^{\prime\prime}$ is $1$, so the degree is $1$.
10. $y^{\prime\prime} + 2y^{\prime} + \sin y = 0$
Answer: The highest-order derivative is $y^{\prime\prime}$, so the order is $2$. The term $\sin y$ involves the dependent variable $y$, not a derivative, so the equation is still a polynomial in its derivatives; the highest power of $y^{\prime\prime}$ is $1$, so the degree is $1$.
11. The degree of the differential equation $\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) + 1 = 0$ is
(A) $3$ (B) $2$ (C) $1$ (D) not defined
Answer: The correct option is (D) not defined. Because the term $\sin\left(\dfrac{dy}{dx}\right)$ appears, the equation is not a polynomial in its derivatives, so its degree cannot be defined.
12. The order of the differential equation $2x^2\dfrac{d^2y}{dx^2} – 3\dfrac{dy}{dx} + y = 0$ is
(A) $2$ (B) $1$ (C) $0$ (D) not defined
Answer: The correct option is (A) $2$. The highest-order derivative in the equation is $\dfrac{d^2y}{dx^2}$, which is of order $2$.
Exercise 9.2
In each of the Exercises 1 to 10, verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.
1. $y = e^x + 1$ : $y^{\prime\prime} – y^{\prime} = 0$
Answer: From $y = e^x + 1$ we get $y^{\prime} = e^x$ and $y^{\prime\prime} = e^x$. Substituting, $y^{\prime\prime} – y^{\prime} = e^x – e^x = 0$. Hence the given function is a solution.
2. $y = x^2 + 2x + C$ : $y^{\prime} – 2x – 2 = 0$
Answer: Differentiating, $y^{\prime} = 2x + 2$. Substituting, $y^{\prime} – 2x – 2 = (2x + 2) – 2x – 2 = 0$. Hence the given function is a solution.
3. $y = \cos x + C$ : $y^{\prime} + \sin x = 0$
Answer: Differentiating, $y^{\prime} = -\sin x$. Substituting, $y^{\prime} + \sin x = -\sin x + \sin x = 0$. Hence the given function is a solution.
4. $y = \sqrt{1 + x^2}$ : $y^{\prime} = \dfrac{xy}{1 + x^2}$
Answer: Differentiating, $y^{\prime} = \dfrac{1}{2\sqrt{1 + x^2}} \cdot 2x = \dfrac{x}{\sqrt{1 + x^2}}$. The right-hand side is $\dfrac{xy}{1 + x^2} = \dfrac{x\sqrt{1 + x^2}}{1 + x^2} = \dfrac{x}{\sqrt{1 + x^2}}$. Since both sides are equal, the given function is a solution.
5. $y = Ax$ : $xy^{\prime} = y$ $(x \neq 0)$
Answer: Differentiating, $y^{\prime} = A$. Then $xy^{\prime} = xA = Ax = y$. Hence the given function is a solution.
6. $y = x\sin x$ : $xy^{\prime} = y + x\sqrt{x^2 – y^2}$ $(x \neq 0$ and $x > y$ or $x < -y)$
Answer: Differentiating, $y^{\prime} = \sin x + x\cos x$, so $xy^{\prime} = x\sin x + x^2\cos x$. On the right-hand side, $\sqrt{x^2 – y^2} = \sqrt{x^2 – x^2\sin^2 x} = |x\cos x| = x\cos x$ (in the stated region where $\cos x > 0$). Then $y + x\sqrt{x^2 – y^2} = x\sin x + x \cdot x\cos x = x\sin x + x^2\cos x$. Both sides are equal, so the given function is a solution.
7. $xy = \log y + C$ : $y^{\prime} = \dfrac{y^2}{1 – xy}$ $(xy \neq 1)$
Answer: Differentiating $xy = \log y + C$ with respect to $x$, $y + xy^{\prime} = \dfrac{1}{y}\,y^{\prime}$, so $y = y^{\prime}\left(\dfrac{1}{y} – x\right) = y^{\prime}\cdot\dfrac{1 – xy}{y}$. Therefore $y^{\prime} = \dfrac{y \cdot y}{1 – xy} = \dfrac{y^2}{1 – xy}$. Hence the given function is a solution.
8. $y – \cos y = x$ : $(y\sin y + \cos y + x)\,y^{\prime} = y$
Answer: Differentiating $y – \cos y = x$, we get $y^{\prime} + \sin y \cdot y^{\prime} = 1$, so $y^{\prime}(1 + \sin y) = 1$, giving $y^{\prime} = \dfrac{1}{1 + \sin y}$. Using $x = y – \cos y$, the bracket becomes $y\sin y + \cos y + x = y\sin y + \cos y + (y – \cos y) = y\sin y + y = y(1 + \sin y)$. Hence $(y\sin y + \cos y + x)y^{\prime} = y(1 + \sin y)\cdot\dfrac{1}{1 + \sin y} = y$. So the given function is a solution.
9. $x + y = \tan^{-1} y$ : $y^2 y^{\prime} + y^2 + 1 = 0$
Answer: Differentiating $x + y = \tan^{-1} y$, we get $1 + y^{\prime} = \dfrac{y^{\prime}}{1 + y^2}$, so $y^{\prime}\left(1 – \dfrac{1}{1 + y^2}\right) = -1$, that is $y^{\prime}\cdot\dfrac{y^2}{1 + y^2} = -1$, giving $y^{\prime} = -\dfrac{1 + y^2}{y^2}$. Then $y^2 y^{\prime} = -(1 + y^2)$, so $y^2 y^{\prime} + y^2 + 1 = -(1 + y^2) + y^2 + 1 = 0$. Hence the given function is a solution.
10. $y = \sqrt{a^2 – x^2}$, $x \in (-a, a)$ : $x + y\dfrac{dy}{dx} = 0$ $(y \neq 0)$
Answer: Differentiating, $\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{a^2 – x^2}} = \dfrac{-x}{\sqrt{a^2 – x^2}} = -\dfrac{x}{y}$. Substituting, $x + y\dfrac{dy}{dx} = x + y\left(-\dfrac{x}{y}\right) = x – x = 0$. Hence the given function is a solution.
11. The number of arbitrary constants in the general solution of a differential equation of fourth order is
(A) $0$ (B) $2$ (C) $3$ (D) $4$
Answer: The correct option is (D) $4$. The general solution of a differential equation contains as many arbitrary constants as the order of the equation, so a fourth-order equation has $4$ arbitrary constants.
12. The number of arbitrary constants in the particular solution of a differential equation of third order is
(A) $3$ (B) $2$ (C) $1$ (D) $0$
Answer: The correct option is (D) $0$. A particular solution is obtained by assigning definite values to all the arbitrary constants, so it contains no arbitrary constant, whatever the order.
Exercise 9.3
For each of the differential equations in Exercises 1 to 10, find the general solution.
1. $\dfrac{dy}{dx} = \dfrac{1 – \cos x}{1 + \cos x}$
Answer: Using $1 – \cos x = 2\sin^2\dfrac{x}{2}$ and $1 + \cos x = 2\cos^2\dfrac{x}{2}$, we get $\dfrac{dy}{dx} = \tan^2\dfrac{x}{2} = \sec^2\dfrac{x}{2} – 1$. Then $dy = \left(\sec^2\dfrac{x}{2} – 1\right)dx$. Integrating, $y = 2\tan\dfrac{x}{2} – x + C$.
2. $\dfrac{dy}{dx} = \sqrt{4 – y^2}$ $(-2 < y < 2)$
Answer: Separating the variables, $\dfrac{dy}{\sqrt{4 – y^2}} = dx$. Integrating, $\sin^{-1}\dfrac{y}{2} = x + C$, that is $y = 2\sin(x + C)$.
3. $\dfrac{dy}{dx} + y = 1$ $(y \neq 1)$
Answer: Writing $\dfrac{dy}{dx} = 1 – y$ and separating, $\dfrac{dy}{1 – y} = dx$. Integrating, $-\log|1 – y| = x + C_1$, so $\log|1 – y| = -x – C_1$ and $1 – y = Ce^{-x}$. Hence $y = 1 – Ce^{-x}$.
4. $\sec^2 x\,\tan y\,dx + \sec^2 y\,\tan x\,dy = 0$
Answer: Dividing throughout by $\tan x\,\tan y$, $\dfrac{\sec^2 x}{\tan x}\,dx + \dfrac{\sec^2 y}{\tan y}\,dy = 0$. Integrating each term, $\log|\tan x| + \log|\tan y| = \log C$. Hence $\tan x\,\tan y = C$.
5. $(e^x + e^{-x})\,dy – (e^x – e^{-x})\,dx = 0$
Answer: Separating, $dy = \dfrac{e^x – e^{-x}}{e^x + e^{-x}}\,dx$. Since the numerator is the derivative of the denominator, integrating gives $y = \log(e^x + e^{-x}) + C$.
6. $\dfrac{dy}{dx} = (1 + x^2)(1 + y^2)$
Answer: Separating, $\dfrac{dy}{1 + y^2} = (1 + x^2)\,dx$. Integrating, $\tan^{-1} y = x + \dfrac{x^3}{3} + C$.
7. $y\log y\,dx – x\,dy = 0$
Answer: Separating, $\dfrac{dy}{y\log y} = \dfrac{dx}{x}$. Putting $\log y = t$ so that $\dfrac{dy}{y} = dt$ on the left, integrating gives $\log|\log y| = \log|x| + \log C$. Hence $\log y = Cx$, that is $y = e^{Cx}$.
8. $x^5\dfrac{dy}{dx} = -y^5$
Answer: Separating, $\dfrac{dy}{y^5} = -\dfrac{dx}{x^5}$. Integrating, $-\dfrac{1}{4y^4} = \dfrac{1}{4x^4} + C_1$. Multiplying by $-4$, $\dfrac{1}{y^4} = -\dfrac{1}{x^4} – 4C_1$, which gives $x^{-4} + y^{-4} = C$ (with $C = -4C_1$).
9. $\dfrac{dy}{dx} = \sin^{-1} x$
Answer: Integrating by parts, $y = \int \sin^{-1} x\,dx = x\sin^{-1} x – \int \dfrac{x}{\sqrt{1 – x^2}}\,dx = x\sin^{-1} x + \sqrt{1 – x^2} + C$.
10. $e^x\tan y\,dx + (1 – e^x)\sec^2 y\,dy = 0$
Answer: Dividing by $\tan y\,(1 – e^x)$, $\dfrac{e^x}{1 – e^x}\,dx + \dfrac{\sec^2 y}{\tan y}\,dy = 0$. Integrating, $-\log|1 – e^x| + \log|\tan y| = \log C$, so $\dfrac{\tan y}{1 – e^x} = C$. Hence $\tan y = C(1 – e^x)$.
For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition.
11. $(x^3 + x^2 + x + 1)\dfrac{dy}{dx} = 2x^2 + x$; $y = 1$ when $x = 0$
Answer: Since $x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1)$, we have $\dfrac{dy}{dx} = \dfrac{2x^2 + x}{(x + 1)(x^2 + 1)}$. Using partial fractions, $\dfrac{2x^2 + x}{(x + 1)(x^2 + 1)} = \dfrac{1/2}{x + 1} + \dfrac{(3/2)x – 1/2}{x^2 + 1}$. Integrating, $y = \dfrac{1}{2}\log|x + 1| + \dfrac{3}{4}\log(x^2 + 1) – \dfrac{1}{2}\tan^{-1} x + C$. At $x = 0$, $y = 1$ gives $C = 1$. Hence $y = \dfrac{1}{2}\log|x + 1| + \dfrac{3}{4}\log(x^2 + 1) – \dfrac{1}{2}\tan^{-1} x + 1$.
12. $x(x^2 – 1)\dfrac{dy}{dx} = 1$; $y = 0$ when $x = 2$
Answer: Here $\dfrac{dy}{dx} = \dfrac{1}{x(x – 1)(x + 1)}$. Using partial fractions, $\dfrac{1}{x(x – 1)(x + 1)} = -\dfrac{1}{x} + \dfrac{1}{2(x – 1)} + \dfrac{1}{2(x + 1)}$. Integrating, $y = -\log|x| + \dfrac{1}{2}\log|x^2 – 1| + C = \dfrac{1}{2}\log\left|\dfrac{x^2 – 1}{x^2}\right| + C$. At $x = 2$, $y = 0$: $0 = \dfrac{1}{2}\log\dfrac{3}{4} + C$, so $C = \dfrac{1}{2}\log\dfrac{4}{3}$. Hence $y = \dfrac{1}{2}\log\left|\dfrac{x^2 – 1}{x^2}\right| + \dfrac{1}{2}\log\dfrac{4}{3}$.
13. $\cos\left(\dfrac{dy}{dx}\right) = a$ $(a \in \mathbb{R})$; $y = 1$ when $x = 0$
Answer: From $\cos\left(\dfrac{dy}{dx}\right) = a$ we get $\dfrac{dy}{dx} = \cos^{-1} a$, a constant. Integrating, $y = x\cos^{-1} a + C$. At $x = 0$, $y = 1$ gives $C = 1$. Hence $y = x\cos^{-1} a + 1$, that is $\dfrac{y – 1}{x} = \cos^{-1} a$, or $\cos\left(\dfrac{y – 1}{x}\right) = a$.
14. $\dfrac{dy}{dx} = y\tan x$; $y = 1$ when $x = 0$
Answer: Separating, $\dfrac{dy}{y} = \tan x\,dx$. Integrating, $\log|y| = \log|\sec x| + C_1$. At $x = 0$, $y = 1$: $\log 1 = \log 1 + C_1$, so $C_1 = 0$. Hence $\log y = \log\sec x$, giving $y = \sec x$.
15. Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y^{\prime} = e^x\sin x$.
Answer: Integrating, $y = \int e^x\sin x\,dx = \dfrac{e^x(\sin x – \cos x)}{2} + C$. At $(0, 0)$: $0 = \dfrac{1(0 – 1)}{2} + C = -\dfrac{1}{2} + C$, so $C = \dfrac{1}{2}$. Hence $y = \dfrac{e^x(\sin x – \cos x)}{2} + \dfrac{1}{2}$, that is $2y – 1 = e^x(\sin x – \cos x)$.
16. For the differential equation $xy\dfrac{dy}{dx} = (x + 2)(y + 2)$, find the solution curve passing through the point $(1, -1)$.
Answer: Separating the variables, $\dfrac{y}{y + 2}\,dy = \dfrac{x + 2}{x}\,dx$, that is $\left(1 – \dfrac{2}{y + 2}\right)dy = \left(1 + \dfrac{2}{x}\right)dx$. Integrating, $y – 2\log|y + 2| = x + 2\log|x| + C$. At $(1, -1)$: $-1 – 2\log 1 = 1 + 2\log 1 + C$, so $C = -2$. Thus $y – x + 2 = 2\log|x| + 2\log|y + 2| = \log\{x^2(y + 2)^2\}$. Hence the required curve is $\log\{x^2(y + 2)^2\} = y – x + 2$.
17. Find the equation of a curve passing through the point $(0, -2)$ given that at any point $(x, y)$ on the curve, the product of the slope of its tangent and the $y$ coordinate of the point is equal to the $x$ coordinate of the point.
Answer: The condition gives $y\dfrac{dy}{dx} = x$, so $y\,dy = x\,dx$. Integrating, $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C_1$, that is $y^2 – x^2 = C$. At $(0, -2)$: $4 – 0 = C$, so $C = 4$. Hence the required curve is $y^2 – x^2 = 4$.
18. At any point $(x, y)$ of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point $(-4, -3)$. Find the equation of the curve given that it passes through $(-2, 1)$.
Answer: The slope of the segment joining $(x, y)$ to $(-4, -3)$ is $\dfrac{y + 3}{x + 4}$, so $\dfrac{dy}{dx} = \dfrac{2(y + 3)}{x + 4}$. Separating, $\dfrac{dy}{y + 3} = \dfrac{2\,dx}{x + 4}$. Integrating, $\log|y + 3| = 2\log|x + 4| + \log C$, so $y + 3 = C(x + 4)^2$. At $(-2, 1)$: $4 = C(2)^2 = 4C$, so $C = 1$. Hence the required curve is $y + 3 = (x + 4)^2$.
19. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units, find the radius of the balloon after $t$ seconds.
Answer: Let $V = \dfrac{4}{3}\pi r^3$ be the volume. Since it changes at a constant rate, $\dfrac{dV}{dt} = k$, so $\dfrac{4}{3}\pi r^3 = kt + C$. At $t = 0$, $r = 3$: $\dfrac{4}{3}\pi(27) = 36\pi = C$. At $t = 3$, $r = 6$: $\dfrac{4}{3}\pi(216) = 288\pi = 3k + 36\pi$, so $k = 84\pi$. Thus $\dfrac{4}{3}\pi r^3 = 84\pi t + 36\pi$, giving $r^3 = 63t + 27$. Hence $r = (63t + 27)^{1/3}$.
20. In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if Rs $100$ doubles itself in $10$ years $(\log_e 2 = 0.6931)$.
Answer: Let $P$ be the principal. Then $\dfrac{dP}{dt} = \dfrac{r}{100}P$, which gives $P = P_0\,e^{rt/100}$. Here $P_0 = 100$ and $P = 200$ at $t = 10$, so $200 = 100\,e^{10r/100} = 100\,e^{r/10}$. Thus $e^{r/10} = 2$, giving $\dfrac{r}{10} = \log_e 2 = 0.6931$. Hence $r = 6.931\%$, that is about $6.93\%$ per year.
21. In a bank, principal increases continuously at the rate of $5\%$ per year. An amount of Rs $1000$ is deposited with this bank. How much will it be worth after $10$ years $(e^{0.5} = 1.648)$?
Answer: As above, $\dfrac{dP}{dt} = \dfrac{5}{100}P = \dfrac{P}{20}$, so $P = 1000\,e^{0.05t}$. At $t = 10$, $P = 1000\,e^{0.5} = 1000(1.648) = $ Rs $1648$. Hence the amount will be worth about Rs $1648$ after $10$ years.
22. In a culture, the bacteria count is $1{,}00{,}000$. The number is increased by $10\%$ in $2$ hours. In how many hours will the count reach $2{,}00{,}000$, if the rate of growth of bacteria is proportional to the number present?
Answer: Let $y$ be the count at time $t$. Then $\dfrac{dy}{dt} = ky$, so $y = y_0\,e^{kt}$ with $y_0 = 1{,}00{,}000$. After $2$ hours, $y = 1{,}10{,}000$, so $1.1 = e^{2k}$, giving $k = \dfrac{1}{2}\log 1.1$. We need $y = 2{,}00{,}000$, that is $e^{kt} = 2$, so $kt = \log 2$ and $t = \dfrac{\log 2}{k} = \dfrac{2\log 2}{\log 1.1}$. Numerically $t \approx 14.55$ hours.
23. The general solution of the differential equation $\dfrac{dy}{dx} = e^{x + y}$ is
(A) $e^x + e^{-y} = C$ (B) $e^x + e^y = C$ (C) $e^{-x} + e^y = C$ (D) $e^{-x} + e^{-y} = C$
Answer: The correct option is (A). Writing $\dfrac{dy}{dx} = e^x\,e^y$ and separating, $e^{-y}\,dy = e^x\,dx$. Integrating, $-e^{-y} = e^x + C_1$, that is $e^x + e^{-y} = C$.
Exercise 9.4
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.
1. $(x^2 + xy)\,dy = (x^2 + y^2)\,dx$
Answer: Here $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{x^2 + xy}$; replacing $x, y$ by $\lambda x, \lambda y$ leaves the right side unchanged, so it is homogeneous of degree zero. Put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Then $v + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{1 + v}$, giving $x\dfrac{dv}{dx} = \dfrac{1 + v^2}{1 + v} – v = \dfrac{1 – v}{1 + v}$. Separating, $\dfrac{1 + v}{1 – v}\,dv = \dfrac{dx}{x}$, that is $\left(-1 + \dfrac{2}{1 – v}\right)dv = \dfrac{dx}{x}$. Integrating, $-v – 2\log|1 – v| = \log|x| + C$. Putting $v = \dfrac{y}{x}$ and simplifying gives $\log\left|\dfrac{x}{(x – y)^2}\right| = \dfrac{y}{x} + C$.
2. $y^{\prime} = \dfrac{x + y}{x}$
Answer: Here $\dfrac{dy}{dx} = 1 + \dfrac{y}{x}$, which is homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = 1 + v$, so $x\dfrac{dv}{dx} = 1$, giving $dv = \dfrac{dx}{x}$. Integrating, $v = \log|x| + C$. Hence $\dfrac{y}{x} = \log|x| + C$, that is $y = x\log|x| + Cx$.
3. $(x – y)\,dy – (x + y)\,dx = 0$
Answer: Here $\dfrac{dy}{dx} = \dfrac{x + y}{x – y}$, homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{1 + v}{1 – v}$, so $x\dfrac{dv}{dx} = \dfrac{1 + v^2}{1 – v}$. Separating, $\dfrac{1 – v}{1 + v^2}\,dv = \dfrac{dx}{x}$, that is $\left(\dfrac{1}{1 + v^2} – \dfrac{v}{1 + v^2}\right)dv = \dfrac{dx}{x}$. Integrating, $\tan^{-1} v – \dfrac{1}{2}\log(1 + v^2) = \log|x| + C$. Putting $v = \dfrac{y}{x}$ and simplifying, $\tan^{-1}\dfrac{y}{x} = \dfrac{1}{2}\log(x^2 + y^2) + C$.
4. $(x^2 – y^2)\,dx + 2xy\,dy = 0$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y^2 – x^2}{2xy}$, homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{v^2 – 1}{2v}$, so $x\dfrac{dv}{dx} = -\dfrac{1 + v^2}{2v}$. Separating, $\dfrac{2v}{1 + v^2}\,dv = -\dfrac{dx}{x}$. Integrating, $\log(1 + v^2) = -\log|x| + \log C$, so $x(1 + v^2) = C$. Putting $v = \dfrac{y}{x}$, $x\left(1 + \dfrac{y^2}{x^2}\right) = C$, that is $x^2 + y^2 = Cx$.
5. $x^2\dfrac{dy}{dx} = x^2 – 2y^2 + xy$
Answer: Here $\dfrac{dy}{dx} = 1 – 2\left(\dfrac{y}{x}\right)^2 + \dfrac{y}{x}$, homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = 1 – 2v^2 + v$, so $x\dfrac{dv}{dx} = 1 – 2v^2$. Separating, $\dfrac{dv}{1 – 2v^2} = \dfrac{dx}{x}$. Integrating, $\dfrac{1}{2\sqrt{2}}\log\left|\dfrac{1 + \sqrt{2}\,v}{1 – \sqrt{2}\,v}\right| = \log|x| + C$. Putting $v = \dfrac{y}{x}$, $\dfrac{1}{2\sqrt{2}}\log\left|\dfrac{x + \sqrt{2}\,y}{x – \sqrt{2}\,y}\right| = \log|x| + C$.
6. $x\,dy – y\,dx = \sqrt{x^2 + y^2}\,dx$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y + \sqrt{x^2 + y^2}}{x}$, homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = v + \sqrt{1 + v^2}$, so $x\dfrac{dv}{dx} = \sqrt{1 + v^2}$. Separating, $\dfrac{dv}{\sqrt{1 + v^2}} = \dfrac{dx}{x}$. Integrating, $\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log C$, so $v + \sqrt{1 + v^2} = Cx$. Putting $v = \dfrac{y}{x}$, $\dfrac{y}{x} + \sqrt{1 + \dfrac{y^2}{x^2}} = Cx$, that is $y + \sqrt{x^2 + y^2} = Cx^2$.
7. $\left\{x\cos\dfrac{y}{x} + y\sin\dfrac{y}{x}\right\}y\,dx = \left\{y\sin\dfrac{y}{x} – x\cos\dfrac{y}{x}\right\}x\,dy$
Answer: Writing $\dfrac{dy}{dx}$ and putting $y = vx$, the equation reduces to $v + x\dfrac{dv}{dx} = \dfrac{v(\cos v + v\sin v)}{v\sin v – \cos v}$, so $x\dfrac{dv}{dx} = \dfrac{2v\cos v}{v\sin v – \cos v}$. Separating, $\dfrac{v\sin v – \cos v}{v\cos v}\,dv = \dfrac{2\,dx}{x}$, that is $\left(\tan v – \dfrac{1}{v}\right)dv = \dfrac{2\,dx}{x}$. Integrating, $-\log|\cos v| – \log|v| = 2\log|x| + \log C$, so $\dfrac{1}{v\cos v} = Cx^2$. Putting $v = \dfrac{y}{x}$, $xy\cos\dfrac{y}{x} = C$.
8. $x\dfrac{dy}{dx} – y + x\sin\dfrac{y}{x} = 0$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y}{x} – \sin\dfrac{y}{x}$, homogeneous of degree zero. Put $y = vx$: $v + x\dfrac{dv}{dx} = v – \sin v$, so $x\dfrac{dv}{dx} = -\sin v$. Separating, $\dfrac{dv}{\sin v} = -\dfrac{dx}{x}$. Integrating, $\log\left|\tan\dfrac{v}{2}\right| = -\log|x| + \log C$. Putting $v = \dfrac{y}{x}$, $\tan\dfrac{y}{2x} = \dfrac{C}{x}$, that is $x\tan\dfrac{y}{2x} = C$.
9. $y\,dx + x\log\dfrac{y}{x}\,dy – 2x\,dy = 0$
Answer: Writing $y\dfrac{dx}{dy} = x\left(2 – \log\dfrac{y}{x}\right)$, this is homogeneous. Put $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ and $\log\dfrac{y}{x} = -\log v$. Then $v + y\dfrac{dv}{dy} = v(2 + \log v)$, so $y\dfrac{dv}{dy} = v(1 + \log v)$. Separating, $\dfrac{dv}{v(1 + \log v)} = \dfrac{dy}{y}$. Putting $1 + \log v = t$, integrating gives $\log|1 + \log v| = \log|y| + \log C$, so $1 + \log v = Cy$. Putting $v = \dfrac{x}{y}$, $1 + \log\dfrac{x}{y} = Cy$.
10. $\left(1 + e^{x/y}\right)dx + e^{x/y}\left(1 – \dfrac{x}{y}\right)dy = 0$
Answer: Here $\dfrac{dx}{dy} = -\dfrac{e^{x/y}\left(1 – \frac{x}{y}\right)}{1 + e^{x/y}}$, homogeneous of degree zero. Put $x = vy$, so $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$ and $e^{x/y} = e^v$. Then $v + y\dfrac{dv}{dy} = -\dfrac{e^v(1 – v)}{1 + e^v}$, so $y\dfrac{dv}{dy} = -\dfrac{v + e^v}{1 + e^v}$. Separating, $\dfrac{1 + e^v}{v + e^v}\,dv = -\dfrac{dy}{y}$. Since the numerator is the derivative of $v + e^v$, integrating gives $\log|v + e^v| = -\log|y| + \log C$, so $y(v + e^v) = C$. Putting $v = \dfrac{x}{y}$, $x + y\,e^{x/y} = C$.
For each of the differential equations in Exercises 11 to 15, find the particular solution satisfying the given condition.
11. $(x + y)\,dy + (x – y)\,dx = 0$; $y = 1$ when $x = 1$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y – x}{x + y}$. Put $y = vx$: $v + x\dfrac{dv}{dx} = \dfrac{v – 1}{1 + v}$, so $x\dfrac{dv}{dx} = -\dfrac{1 + v^2}{1 + v}$. Separating, $\dfrac{1 + v}{1 + v^2}\,dv = -\dfrac{dx}{x}$. Integrating, $\tan^{-1} v + \dfrac{1}{2}\log(1 + v^2) = -\log|x| + C$. Putting $v = \dfrac{y}{x}$ and simplifying, $\tan^{-1}\dfrac{y}{x} + \dfrac{1}{2}\log(x^2 + y^2) = C$. At $(1, 1)$: $\dfrac{\pi}{4} + \dfrac{1}{2}\log 2 = C$. Hence $\dfrac{1}{2}\log(x^2 + y^2) + \tan^{-1}\dfrac{y}{x} = \dfrac{\pi}{4} + \dfrac{1}{2}\log 2$, or $\log(x^2 + y^2) + 2\tan^{-1}\dfrac{y}{x} = \dfrac{\pi}{2} + \log 2$.
12. $x^2\,dy + (xy + y^2)\,dx = 0$; $y = 1$ when $x = 1$
Answer: Here $\dfrac{dy}{dx} = -\dfrac{xy + y^2}{x^2}$. Put $y = vx$: $v + x\dfrac{dv}{dx} = -(v + v^2)$, so $x\dfrac{dv}{dx} = -v(v + 2)$. Separating, $\dfrac{dv}{v(v + 2)} = -\dfrac{dx}{x}$, that is $\dfrac{1}{2}\left(\dfrac{1}{v} – \dfrac{1}{v + 2}\right)dv = -\dfrac{dx}{x}$. Integrating, $\dfrac{1}{2}\log\left|\dfrac{v}{v + 2}\right| = -\log|x| + C_1$, so $\dfrac{v}{v + 2} = \dfrac{A}{x^2}$. Putting $v = \dfrac{y}{x}$, $\dfrac{y}{y + 2x} = \dfrac{A}{x^2}$. At $(1, 1)$: $\dfrac{1}{3} = A$. Hence $\dfrac{y}{y + 2x} = \dfrac{1}{3x^2}$, which gives $3x^2 y = y + 2x$, that is $y = \dfrac{2x}{3x^2 – 1}$.
13. $\left[x\sin^2\dfrac{y}{x} – y\right]dx + x\,dy = 0$; $y = \dfrac{\pi}{4}$ when $x = 1$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y}{x} – \sin^2\dfrac{y}{x}$. Put $y = vx$: $v + x\dfrac{dv}{dx} = v – \sin^2 v$, so $x\dfrac{dv}{dx} = -\sin^2 v$. Separating, $\dfrac{dv}{\sin^2 v} = -\dfrac{dx}{x}$, that is $\text{cosec}^2 v\,dv = -\dfrac{dx}{x}$. Integrating, $-\cot v = -\log|x| + C$, so $\cot v = \log|x| + C_1$. At $x = 1$, $v = \dfrac{y}{x} = \dfrac{\pi}{4}$: $\cot\dfrac{\pi}{4} = 1 = 0 + C_1$, so $C_1 = 1$. Hence $\cot\dfrac{y}{x} = \log|x| + 1 = \log|ex|$.
14. $\dfrac{dy}{dx} – \dfrac{y}{x} + \text{cosec}\dfrac{y}{x} = 0$; $y = 0$ when $x = 1$
Answer: Here $\dfrac{dy}{dx} = \dfrac{y}{x} – \text{cosec}\dfrac{y}{x}$. Put $y = vx$: $v + x\dfrac{dv}{dx} = v – \text{cosec}\,v$, so $x\dfrac{dv}{dx} = -\text{cosec}\,v$. Separating, $\sin v\,dv = -\dfrac{dx}{x}$. Integrating, $-\cos v = -\log|x| + C$, so $\cos v = \log|x| + C_1$. At $x = 1$, $v = 0$: $\cos 0 = 1 = 0 + C_1$, so $C_1 = 1$. Hence $\cos\dfrac{y}{x} = \log|x| + 1 = \log|ex|$.
15. $2xy + y^2 – 2x^2\dfrac{dy}{dx} = 0$; $y = 2$ when $x = 1$
Answer: Here $\dfrac{dy}{dx} = \dfrac{2xy + y^2}{2x^2}$. Put $y = vx$: $v + x\dfrac{dv}{dx} = v + \dfrac{v^2}{2}$, so $x\dfrac{dv}{dx} = \dfrac{v^2}{2}$. Separating, $\dfrac{dv}{v^2} = \dfrac{dx}{2x}$. Integrating, $-\dfrac{1}{v} = \dfrac{1}{2}\log|x| + C$. Putting $v = \dfrac{y}{x}$, $-\dfrac{x}{y} = \dfrac{1}{2}\log|x| + C$. At $(1, 2)$: $-\dfrac{1}{2} = 0 + C$, so $C = -\dfrac{1}{2}$. Then $\dfrac{2x}{y} = 1 – \log|x|$, giving $y = \dfrac{2x}{1 – \log|x|}$.
16. A homogeneous differential equation of the form $\dfrac{dx}{dy} = h\left(\dfrac{x}{y}\right)$ can be solved by making the substitution
(A) $y = vx$ (B) $v = yx$ (C) $x = vy$ (D) $x = v$
Answer: The correct option is (C) $x = vy$. When the equation is expressed with $x$ as a function of $y$ in terms of $\dfrac{x}{y}$, the appropriate substitution is $x = vy$.
17. Which of the following is a homogeneous differential equation?
(A) $(4x + 6y + 5)\,dy – (3y + 2x + 4)\,dx = 0$
(B) $(xy)\,dx – (x^3 + y^3)\,dy = 0$
(C) $(x^3 + 2y^2)\,dx + 2xy\,dy = 0$
(D) $y^2\,dx + (x^2 – xy – y^2)\,dy = 0$
Answer: The correct option is (D). In option (D) every term — $y^2$ and $x^2 – xy – y^2$ — is homogeneous of degree $2$, so the equation is homogeneous. In (A) the constants make it non-homogeneous, while in (B) and (C) the terms are of different degrees.
Exercise 9.5
For each of the differential equations given in Exercises 1 to 12, find the general solution.
1. $\dfrac{dy}{dx} + 2y = \sin x$
Answer: Here $P = 2$, $Q = \sin x$, so $\text{I.F.} = e^{\int 2\,dx} = e^{2x}$. The solution is $y\,e^{2x} = \int e^{2x}\sin x\,dx = \dfrac{e^{2x}(2\sin x – \cos x)}{5} + C$. Hence $y = \dfrac{2\sin x – \cos x}{5} + Ce^{-2x}$.
2. $\dfrac{dy}{dx} + 3y = e^{-2x}$
Answer: Here $\text{I.F.} = e^{\int 3\,dx} = e^{3x}$. The solution is $y\,e^{3x} = \int e^{3x}\,e^{-2x}\,dx = \int e^x\,dx = e^x + C$. Hence $y = e^{-2x} + Ce^{-3x}$.
3. $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$
Answer: Here $P = \dfrac{1}{x}$, so $\text{I.F.} = e^{\int dx/x} = x$. The solution is $y\,x = \int x \cdot x^2\,dx = \dfrac{x^4}{4} + C$. Hence $y = \dfrac{x^3}{4} + \dfrac{C}{x}$.
4. $\dfrac{dy}{dx} + (\sec x)\,y = \tan x$ $\left(0 \le x < \dfrac{\pi}{2}\right)$
Answer: Here $\text{I.F.} = e^{\int \sec x\,dx} = e^{\log|\sec x + \tan x|} = \sec x + \tan x$. The solution is $y(\sec x + \tan x) = \int \tan x(\sec x + \tan x)\,dx = \int (\sec x\tan x + \sec^2 x – 1)\,dx = \sec x + \tan x – x + C$. Hence $y(\sec x + \tan x) = \sec x + \tan x – x + C$.
5. $\cos^2 x\dfrac{dy}{dx} + y = \tan x$ $\left(0 \le x < \dfrac{\pi}{2}\right)$
Answer: Dividing by $\cos^2 x$, $\dfrac{dy}{dx} + (\sec^2 x)\,y = \tan x\sec^2 x$, so $\text{I.F.} = e^{\int \sec^2 x\,dx} = e^{\tan x}$. The solution is $y\,e^{\tan x} = \int \tan x\sec^2 x\,e^{\tan x}\,dx$. Putting $t = \tan x$, $\int t\,e^t\,dt = e^t(t – 1)$, so $y\,e^{\tan x} = e^{\tan x}(\tan x – 1) + C$. Hence $y = (\tan x – 1) + Ce^{-\tan x}$.
6. $x\dfrac{dy}{dx} + 2y = x^2\log x$
Answer: Dividing by $x$, $\dfrac{dy}{dx} + \dfrac{2}{x}y = x\log x$, so $\text{I.F.} = e^{\int 2\,dx/x} = x^2$. The solution is $y\,x^2 = \int x^2 \cdot x\log x\,dx = \int x^3\log x\,dx = \dfrac{x^4}{4}\log x – \dfrac{x^4}{16} + C$. Hence $y = \dfrac{x^2}{4}\log x – \dfrac{x^2}{16} + \dfrac{C}{x^2}$.
7. $x\log x\dfrac{dy}{dx} + y = \dfrac{2}{x}\log x$
Answer: Dividing by $x\log x$, $\dfrac{dy}{dx} + \dfrac{y}{x\log x} = \dfrac{2}{x^2}$, so $\text{I.F.} = e^{\int dx/(x\log x)} = e^{\log(\log x)} = \log x$. The solution is $y\log x = \int \dfrac{2}{x^2}\log x\,dx = -\dfrac{2}{x}(1 + \log x) + C$. Hence $y\log x = -\dfrac{2}{x}(1 + \log x) + C$.
8. $(1 + x^2)\,dy + 2xy\,dx = \cot x\,dx$ $(x \neq 0)$
Answer: Dividing by $1 + x^2$, $\dfrac{dy}{dx} + \dfrac{2x}{1 + x^2}\,y = \dfrac{\cot x}{1 + x^2}$, so $\text{I.F.} = e^{\int 2x\,dx/(1 + x^2)} = 1 + x^2$. The solution is $y(1 + x^2) = \int \dfrac{\cot x}{1 + x^2}(1 + x^2)\,dx = \int \cot x\,dx = \log|\sin x| + C$. Hence $y(1 + x^2) = \log|\sin x| + C$.
9. $x\dfrac{dy}{dx} + y – x + xy\cot x = 0$ $(x \neq 0)$
Answer: Rearranging, $\dfrac{dy}{dx} + \left(\dfrac{1}{x} + \cot x\right)y = 1$, so $\text{I.F.} = e^{\int (1/x + \cot x)\,dx} = e^{\log x + \log\sin x} = x\sin x$. The solution is $y \cdot x\sin x = \int x\sin x\,dx = \sin x – x\cos x + C$. Hence $y = \dfrac{1}{x} – \cot x + \dfrac{C}{x\sin x}$.
10. $(x + y)\dfrac{dy}{dx} = 1$
Answer: Writing it as $\dfrac{dx}{dy} = x + y$, that is $\dfrac{dx}{dy} – x = y$, this is linear in $x$ with $\text{I.F.} = e^{-y}$. The solution is $x\,e^{-y} = \int y\,e^{-y}\,dy = -(y + 1)e^{-y} + C$. Hence $x = -(y + 1) + Ce^y$, that is $x = Ce^y – y – 1$.
11. $y\,dx + (x – y^2)\,dy = 0$
Answer: Writing it as $\dfrac{dx}{dy} + \dfrac{x}{y} = y$, this is linear in $x$ with $\text{I.F.} = e^{\int dy/y} = y$. The solution is $x\,y = \int y \cdot y\,dy = \dfrac{y^3}{3} + C$. Hence $x = \dfrac{y^2}{3} + \dfrac{C}{y}$.
12. $(x + 3y^2)\dfrac{dy}{dx} = y$ $(y > 0)$
Answer: Writing it as $\dfrac{dx}{dy} = \dfrac{x + 3y^2}{y} = \dfrac{x}{y} + 3y$, that is $\dfrac{dx}{dy} – \dfrac{x}{y} = 3y$, this is linear in $x$ with $\text{I.F.} = e^{-\int dy/y} = \dfrac{1}{y}$. The solution is $x \cdot \dfrac{1}{y} = \int 3y \cdot \dfrac{1}{y}\,dy = 3y + C$. Hence $x = 3y^2 + Cy$.
For each of the differential equations given in Exercises 13 to 15, find a particular solution satisfying the given condition.
13. $\dfrac{dy}{dx} + 2y\tan x = \sin x$; $y = 0$ when $x = \dfrac{\pi}{3}$
Answer: Here $\text{I.F.} = e^{\int 2\tan x\,dx} = e^{2\log\sec x} = \sec^2 x$. The solution is $y\sec^2 x = \int \sin x\sec^2 x\,dx = \int \sec x\tan x\,dx = \sec x + C$, so $y = \cos x + C\cos^2 x$. At $x = \dfrac{\pi}{3}$, $y = 0$: $0 = \dfrac{1}{2} + \dfrac{C}{4}$, so $C = -2$. Hence $y = \cos x – 2\cos^2 x$.
14. $(1 + x^2)\dfrac{dy}{dx} + 2xy = \dfrac{1}{1 + x^2}$; $y = 0$ when $x = 1$
Answer: Dividing by $1 + x^2$, $\dfrac{dy}{dx} + \dfrac{2x}{1 + x^2}\,y = \dfrac{1}{(1 + x^2)^2}$, so $\text{I.F.} = 1 + x^2$. The solution is $y(1 + x^2) = \int \dfrac{1}{(1 + x^2)^2}(1 + x^2)\,dx = \int \dfrac{dx}{1 + x^2} = \tan^{-1} x + C$. At $x = 1$, $y = 0$: $0 = \dfrac{\pi}{4} + C$, so $C = -\dfrac{\pi}{4}$. Hence $y(1 + x^2) = \tan^{-1} x – \dfrac{\pi}{4}$.
15. $\dfrac{dy}{dx} – 3y\cot x = \sin 2x$; $y = 2$ when $x = \dfrac{\pi}{2}$
Answer: Here $\text{I.F.} = e^{-3\int \cot x\,dx} = e^{-3\log\sin x} = \dfrac{1}{\sin^3 x}$. The solution is $\dfrac{y}{\sin^3 x} = \int \dfrac{\sin 2x}{\sin^3 x}\,dx = \int \dfrac{2\cos x}{\sin^2 x}\,dx = -\dfrac{2}{\sin x} + C$, so $y = -2\sin^2 x + C\sin^3 x$. At $x = \dfrac{\pi}{2}$, $y = 2$: $2 = -2 + C$, so $C = 4$. Hence $y = 4\sin^3 x – 2\sin^2 x$.
16. Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point $(x, y)$ is equal to the sum of the coordinates of the point.
Answer: The condition gives $\dfrac{dy}{dx} = x + y$, that is $\dfrac{dy}{dx} – y = x$. This is linear with $\text{I.F.} = e^{-x}$, so $y\,e^{-x} = \int x\,e^{-x}\,dx = -(x + 1)e^{-x} + C$, giving $y = -(x + 1) + Ce^x$. At $(0, 0)$: $0 = -1 + C$, so $C = 1$. Hence $y = e^x – x – 1$, that is $x + y + 1 = e^x$.
17. Find the equation of a curve passing through the point $(0, 2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by $5$.
Answer: Taking the slope as $\dfrac{dy}{dx}$, the condition gives $x + y = \dfrac{dy}{dx} + 5$, that is $\dfrac{dy}{dx} – y = x – 5$. This is linear with $\text{I.F.} = e^{-x}$, so $y\,e^{-x} = \int (x – 5)e^{-x}\,dx = -(x – 4)e^{-x} + C$, giving $y = -(x – 4) + Ce^x = 4 – x + Ce^x$. At $(0, 2)$: $2 = 4 + C$, so $C = -2$. Hence $y = 4 – x – 2e^x$.
18. The Integrating Factor of the differential equation $x\dfrac{dy}{dx} – y = 2x^2$ is
(A) $e^{-x}$ (B) $e^{-y}$ (C) $\dfrac{1}{x}$ (D) $x$
Answer: The correct option is (C) $\dfrac{1}{x}$. Writing the equation as $\dfrac{dy}{dx} – \dfrac{y}{x} = 2x$, we have $P = -\dfrac{1}{x}$, so $\text{I.F.} = e^{\int -dx/x} = e^{-\log x} = \dfrac{1}{x}$.
19. The Integrating Factor of the differential equation $(1 – y^2)\dfrac{dx}{dy} + yx = ay$ $(-1 < y < 1)$ is
(A) $\dfrac{1}{y^2 – 1}$ (B) $\dfrac{1}{\sqrt{y^2 – 1}}$ (C) $\dfrac{1}{1 – y^2}$ (D) $\dfrac{1}{\sqrt{1 – y^2}}$
Answer: The correct option is (D) $\dfrac{1}{\sqrt{1 – y^2}}$. Dividing by $1 – y^2$, $\dfrac{dx}{dy} + \dfrac{y}{1 – y^2}\,x = \dfrac{ay}{1 – y^2}$, so $\text{I.F.} = e^{\int y\,dy/(1 – y^2)} = e^{-\frac{1}{2}\log(1 – y^2)} = (1 – y^2)^{-1/2} = \dfrac{1}{\sqrt{1 – y^2}}$.
Miscellaneous Exercise on Chapter 9
1. For each of the differential equations given below, indicate its order and degree (if defined).
(i) $\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 – 6y = \log x$
Answer: The highest-order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. The equation is a polynomial in its derivatives and the highest power of $\dfrac{d^2y}{dx^2}$ is $1$, so the degree is $1$.
(ii) $\left(\dfrac{dy}{dx}\right)^3 – 4\left(\dfrac{dy}{dx}\right)^2 + 7y = \sin x$
Answer: The highest-order derivative is $\dfrac{dy}{dx}$, so the order is $1$. The highest power of $\dfrac{dy}{dx}$ is $3$, so the degree is $3$.
(iii) $\dfrac{d^4y}{dx^4} – \sin\left(\dfrac{d^3y}{dx^3}\right) = 0$
Answer: The highest-order derivative is $\dfrac{d^4y}{dx^4}$, so the order is $4$. Because of the term $\sin\left(\dfrac{d^3y}{dx^3}\right)$ the equation is not a polynomial in its derivatives, so the degree is not defined.
2. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
(i) $xy = ae^x + be^{-x} + x^2$ : $x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} – xy + x^2 – 2 = 0$
Answer: Differentiating $xy = ae^x + be^{-x} + x^2$, $y + xy^{\prime} = ae^x – be^{-x} + 2x$. Differentiating again, $2y^{\prime} + xy^{\prime\prime} = ae^x + be^{-x} + 2$. Since $ae^x + be^{-x} = xy – x^2$ from the given relation, $xy^{\prime\prime} + 2y^{\prime} = xy – x^2 + 2$, that is $x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} – xy + x^2 – 2 = 0$. Verified.
(ii) $y = e^x(a\cos x + b\sin x)$ : $\dfrac{d^2y}{dx^2} – 2\dfrac{dy}{dx} + 2y = 0$
Answer: Here $y^{\prime} = e^x(a\cos x + b\sin x) + e^x(-a\sin x + b\cos x) = y + e^x(b\cos x – a\sin x)$. Differentiating again, $y^{\prime\prime} = y^{\prime} + e^x(b\cos x – a\sin x) + e^x(-b\sin x – a\cos x) = y^{\prime} + (y^{\prime} – y) – y = 2y^{\prime} – 2y$. Hence $y^{\prime\prime} – 2y^{\prime} + 2y = 0$. Verified.
(iii) $y = x\sin 3x$ : $\dfrac{d^2y}{dx^2} + 9y – 6\cos 3x = 0$
Answer: Here $y^{\prime} = \sin 3x + 3x\cos 3x$ and $y^{\prime\prime} = 3\cos 3x + 3\cos 3x – 9x\sin 3x = 6\cos 3x – 9x\sin 3x$. Then $y^{\prime\prime} + 9y – 6\cos 3x = (6\cos 3x – 9x\sin 3x) + 9x\sin 3x – 6\cos 3x = 0$. Verified.
(iv) $x^2 = 2y^2\log y$ : $(x^2 + y^2)\dfrac{dy}{dx} – xy = 0$
Answer: Differentiating $x^2 = 2y^2\log y$, $2x = 2\left(2y\log y + y\right)y^{\prime} = 2y(2\log y + 1)y^{\prime}$, so $y^{\prime} = \dfrac{x}{y(2\log y + 1)}$. From the given relation $2\log y = \dfrac{x^2}{y^2}$, so $2\log y + 1 = \dfrac{x^2 + y^2}{y^2}$. Hence $y^{\prime} = \dfrac{xy}{x^2 + y^2}$, which gives $(x^2 + y^2)\dfrac{dy}{dx} – xy = 0$. Verified.
3. Prove that $x^2 – y^2 = c(x^2 + y^2)^2$ is the general solution of the differential equation $(x^3 – 3xy^2)\,dx = (y^3 – 3x^2y)\,dy$, where $c$ is a parameter.
Answer: Differentiating $x^2 – y^2 = c(x^2 + y^2)^2$, we get $2x – 2yy^{\prime} = 4c(x^2 + y^2)(x + yy^{\prime})$, so $x – yy^{\prime} = 2c(x^2 + y^2)(x + yy^{\prime})$. From the given relation $c = \dfrac{x^2 – y^2}{(x^2 + y^2)^2}$, so $x – yy^{\prime} = \dfrac{2(x^2 – y^2)(x + yy^{\prime})}{x^2 + y^2}$. Multiplying by $(x^2 + y^2)$ and simplifying, $(x – yy^{\prime})(x^2 + y^2) = 2(x^2 – y^2)(x + yy^{\prime})$. Expanding both sides and collecting terms gives $-x^3 + 3xy^2 + (y^3 – 3x^2y)y^{\prime} = 0$, that is $(y^3 – 3x^2y)\,dy = (x^3 – 3xy^2)\,dx$, which is the given differential equation. Since the relation contains one arbitrary constant $c$ (the order of the equation being one), it is the general solution.
4. Find the general solution of the differential equation $\dfrac{dy}{dx} + \sqrt{\dfrac{1 – y^2}{1 – x^2}} = 0$.
Answer: Writing $\dfrac{dy}{dx} = -\dfrac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}$ and separating, $\dfrac{dy}{\sqrt{1 – y^2}} = -\dfrac{dx}{\sqrt{1 – x^2}}$. Integrating, $\sin^{-1} y = -\sin^{-1} x + C$. Hence $\sin^{-1} x + \sin^{-1} y = C$.
5. Show that the general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{y^2 + y + 1}{x^2 + x + 1} = 0$ is given by $(x + y + 1) = A(1 – x – y – 2xy)$, where $A$ is a parameter.
Answer: Separating, $\dfrac{dy}{y^2 + y + 1} = -\dfrac{dx}{x^2 + x + 1}$. Since $y^2 + y + 1 = \left(y + \tfrac{1}{2}\right)^2 + \tfrac{3}{4}$, integrating gives $\dfrac{2}{\sqrt{3}}\tan^{-1}\dfrac{2y + 1}{\sqrt{3}} = -\dfrac{2}{\sqrt{3}}\tan^{-1}\dfrac{2x + 1}{\sqrt{3}} + C$, so $\tan^{-1}\dfrac{2x + 1}{\sqrt{3}} + \tan^{-1}\dfrac{2y + 1}{\sqrt{3}} = C_1$. Applying $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\dfrac{A + B}{1 – AB}$ with $A = \dfrac{2x + 1}{\sqrt{3}}$, $B = \dfrac{2y + 1}{\sqrt{3}}$ gives $\dfrac{\sqrt{3}(x + y + 1)}{1 – x – y – 2xy} = \tan C_1$, a constant. Hence $(x + y + 1) = A(1 – x – y – 2xy)$, where $A = \dfrac{\tan C_1}{\sqrt{3}}$.
6. Find the equation of the curve passing through the point $\left(0, \dfrac{\pi}{4}\right)$ whose differential equation is $\sin x\cos y\,dx + \cos x\sin y\,dy = 0$.
Answer: Dividing by $\cos x\cos y$, $\tan x\,dx + \tan y\,dy = 0$. Integrating, $\log|\sec x| + \log|\sec y| = \log C$, so $\sec x\sec y = C$, that is $\cos x\cos y = C_1$. At $\left(0, \dfrac{\pi}{4}\right)$: $\cos 0\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$. Hence the required curve is $\cos x\cos y = \dfrac{1}{\sqrt{2}}$.
7. Find the particular solution of the differential equation $(1 + e^{2x})\,dy + (1 + y^2)e^x\,dx = 0$, given that $y = 1$ when $x = 0$.
Answer: Separating, $\dfrac{dy}{1 + y^2} = -\dfrac{e^x\,dx}{1 + e^{2x}}$. Putting $t = e^x$ on the right, $\int \dfrac{dy}{1 + y^2} = -\int \dfrac{dt}{1 + t^2}$, so $\tan^{-1} y = -\tan^{-1}(e^x) + C$, that is $\tan^{-1} y + \tan^{-1}(e^x) = C$. At $(0, 1)$: $\dfrac{\pi}{4} + \dfrac{\pi}{4} = C$, so $C = \dfrac{\pi}{2}$. Hence $\tan^{-1} y + \tan^{-1}(e^x) = \dfrac{\pi}{2}$.
8. Solve the differential equation $y\,e^{x/y}\,dx = \left(x\,e^{x/y} + y^2\right)dy$ $(y \neq 0)$.
Answer: Writing $y\,e^{x/y}\dfrac{dx}{dy} – x\,e^{x/y} = y^2$, we get $e^{x/y}\left(y\dfrac{dx}{dy} – x\right) = y^2$. Dividing by $y^2$, $e^{x/y}\cdot\dfrac{y\frac{dx}{dy} – x}{y^2} = 1$, that is $e^{x/y}\dfrac{d}{dy}\left(\dfrac{x}{y}\right) = 1$. Putting $t = \dfrac{x}{y}$, $e^t\,dt = dy$. Integrating, $e^t = y + C$. Hence $e^{x/y} = y + C$.
9. Find a particular solution of the differential equation $(x – y)(dx + dy) = dx – dy$, given that $y = -1$ when $x = 0$. (Hint: put $x – y = t$.)
Answer: Put $t = x – y$, so $dt = dx – dy$ and $dx + dy = 2\,dx – dt$. The equation becomes $t(2\,dx – dt) = dt$, so $2t\,dx = (1 + t)\,dt$, that is $2\,dx = \left(\dfrac{1}{t} + 1\right)dt$. Integrating, $2x = \log|t| + t + C$, so $2x = \log|x – y| + (x – y) + C$, giving $x + y = \log|x – y| + C$. At $(0, -1)$: $-1 = \log 1 + C$, so $C = -1$. Hence $\log|x – y| = x + y + 1$.
10. Solve the differential equation $\left[\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} – \dfrac{y}{\sqrt{x}}\right]\dfrac{dx}{dy} = 1$ $(x \neq 0)$.
Answer: Rewriting, $\dfrac{dy}{dx} = \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} – \dfrac{y}{\sqrt{x}}$, so $\dfrac{dy}{dx} + \dfrac{y}{\sqrt{x}} = \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}$. This is linear with $P = \dfrac{1}{\sqrt{x}}$, so $\text{I.F.} = e^{\int x^{-1/2}\,dx} = e^{2\sqrt{x}}$. The solution is $y\,e^{2\sqrt{x}} = \int \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}\,e^{2\sqrt{x}}\,dx = \int \dfrac{dx}{\sqrt{x}} = 2\sqrt{x} + C$. Hence $y\,e^{2\sqrt{x}} = 2\sqrt{x} + C$.
11. Find a particular solution of the differential equation $\dfrac{dy}{dx} + y\cot x = 4x\,\text{cosec}\,x$ $(x \neq 0)$, given that $y = 0$ when $x = \dfrac{\pi}{2}$.
Answer: Here $\text{I.F.} = e^{\int \cot x\,dx} = \sin x$. The solution is $y\sin x = \int 4x\,\text{cosec}\,x\sin x\,dx = \int 4x\,dx = 2x^2 + C$. At $x = \dfrac{\pi}{2}$, $y = 0$: $0 = 2 \cdot \dfrac{\pi^2}{4} + C = \dfrac{\pi^2}{2} + C$, so $C = -\dfrac{\pi^2}{2}$. Hence $y\sin x = 2x^2 – \dfrac{\pi^2}{2}$.
12. Find a particular solution of the differential equation $(x + 1)\dfrac{dy}{dx} = 2e^{-y} – 1$, given that $y = 0$ when $x = 0$.
Answer: Separating, $\dfrac{dy}{2e^{-y} – 1} = \dfrac{dx}{x + 1}$. Multiplying the left side by $\dfrac{e^y}{e^y}$, $\dfrac{e^y\,dy}{2 – e^y} = \dfrac{dx}{x + 1}$. Integrating, $-\log|2 – e^y| = \log|x + 1| + C$. At $(0, 0)$: $-\log 1 = \log 1 + C$, so $C = 0$. Then $\log|2 – e^y| = -\log|x + 1|$, giving $(2 – e^y)(x + 1) = 1$, that is $e^y = \dfrac{2x + 1}{x + 1}$, or $y = \log\left|\dfrac{2x + 1}{x + 1}\right|$.
13. The general solution of the differential equation $\dfrac{y\,dx – x\,dy}{y} = 0$ is
(A) $xy = C$ (B) $x = Cy^2$ (C) $y = Cx$ (D) $y = Cx^2$
Answer: The correct option is (C) $y = Cx$. From $y\,dx – x\,dy = 0$ (with $y \neq 0$) we get $\dfrac{dy}{y} = \dfrac{dx}{x}$, so integrating gives $\log|y| = \log|x| + \log C$, that is $y = Cx$.
14. The general solution of a differential equation of the type $\dfrac{dx}{dy} + P_1 x = Q_1$ is
(A) $y\,e^{\int P_1\,dy} = \int\left(Q_1 e^{\int P_1\,dy}\right)dy + C$
(B) $y \cdot e^{\int P_1\,dx} = \int\left(Q_1 e^{\int P_1\,dx}\right)dx + C$
(C) $x\,e^{\int P_1\,dy} = \int\left(Q_1 e^{\int P_1\,dy}\right)dy + C$
(D) $x\,e^{\int P_1\,dx} = \int\left(Q_1 e^{\int P_1\,dx}\right)dx + C$
Answer: The correct option is (C). For the linear equation $\dfrac{dx}{dy} + P_1 x = Q_1$ (with $x$ as the dependent variable), the integrating factor is $e^{\int P_1\,dy}$ and the solution is $x\,e^{\int P_1\,dy} = \int\left(Q_1 e^{\int P_1\,dy}\right)dy + C$.
15. The general solution of the differential equation $e^x\,dy + (y\,e^x + 2x)\,dx = 0$ is
(A) $x\,e^y + x^2 = C$ (B) $x\,e^y + y^2 = C$ (C) $y\,e^x + x^2 = C$ (D) $y\,e^y + x^2 = C$
Answer: The correct option is (C) $y\,e^x + x^2 = C$. Since $e^x\,dy + y\,e^x\,dx = d(y\,e^x)$ and $2x\,dx = d(x^2)$, the equation becomes $d(y\,e^x) + d(x^2) = 0$, so integrating gives $y\,e^x + x^2 = C$.
Additional Questions and Answers
Multiple Choice Questions
1. The order of the differential equation $\dfrac{d^2y}{dx^2} + 3\left(\dfrac{dy}{dx}\right)^3 + y = 0$ is
(A) $1$ (B) $2$ (C) $3$ (D) not defined
Answer: (B) $2$. The highest-order derivative present is $\dfrac{d^2y}{dx^2}$, so the order is $2$.
2. The degree of the differential equation $\left(\dfrac{dy}{dx}\right)^2 + \dfrac{dy}{dx} – \sin^2 y = 0$ is
(A) $1$ (B) $2$ (C) $3$ (D) not defined
Answer: (B) $2$. It is a polynomial in $\dfrac{dy}{dx}$ (the term $\sin^2 y$ involves $y$, not a derivative), and the highest power of $\dfrac{dy}{dx}$ is $2$.
3. The number of arbitrary constants in the general solution of a second-order differential equation is
(A) $0$ (B) $1$ (C) $2$ (D) $3$
Answer: (C) $2$. The general solution contains as many arbitrary constants as the order of the equation.
4. The integrating factor of $\dfrac{dy}{dx} + \dfrac{2y}{x} = x^2$ is
(A) $x$ (B) $x^2$ (C) $\dfrac{1}{x^2}$ (D) $2x$
Answer: (B) $x^2$. Here $P = \dfrac{2}{x}$, so $\text{I.F.} = e^{\int 2\,dx/x} = e^{2\log x} = x^2$.
5. The general solution of $\dfrac{dy}{dx} = e^{x – y}$ is
(A) $e^y = e^x + C$ (B) $e^{-y} + e^x = C$ (C) $e^y + e^x = C$ (D) $e^{-y} = e^x + C$
Answer: (A) $e^y = e^x + C$. Writing $\dfrac{dy}{dx} = e^x e^{-y}$ and separating, $e^y\,dy = e^x\,dx$, so $e^y = e^x + C$.
6. The general solution of $\dfrac{dy}{dx} = \dfrac{y}{x}$ is
(A) $y = Cx$ (B) $y = \dfrac{C}{x}$ (C) $xy = C$ (D) $y = x + C$
Answer: (A) $y = Cx$. Separating, $\dfrac{dy}{y} = \dfrac{dx}{x}$, so $\log y = \log x + \log C$, giving $y = Cx$.
7. Which of the following is a homogeneous function of degree $2$?
(A) $x^2 + xy$ (B) $x^2 + y$ (C) $x + y$ (D) $x^2 + y^2 + x$
Answer: (A) $x^2 + xy$. Replacing $x, y$ by $\lambda x, \lambda y$ gives $\lambda^2(x^2 + xy)$, so it is homogeneous of degree $2$; the other options are not homogeneous.
8. The integrating factor of the differential equation $\dfrac{dy}{dx} + y = e^{-x}$ is
(A) $e^x$ (B) $e^{-x}$ (C) $x$ (D) $e^{2x}$
Answer: (A) $e^x$. Here $P = 1$, so $\text{I.F.} = e^{\int 1\,dx} = e^x$.
Fill in the Blanks
1. The order of a differential equation is the order of the ________ derivative present in it.
Answer: highest-order
2. A differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P$ and $Q$ are functions of $x$ only, is called a first-order ________ differential equation.
Answer: linear
3. To solve a homogeneous differential equation $\dfrac{dy}{dx} = F(x, y)$, we use the substitution ________.
Answer: $y = vx$
4. The integrating factor of the linear differential equation $\dfrac{dy}{dx} + Py = Q$ is ________.
Answer: $e^{\int P\,dx}$
5. A solution of a differential equation containing arbitrary constants equal in number to the order of the equation is called its ________ solution.
Answer: general
True or False
1. The degree of the differential equation $\dfrac{d^2y}{dx^2} + \sin\left(\dfrac{dy}{dx}\right) = 0$ is $1$.
Answer: False. Because of the term $\sin\left(\dfrac{dy}{dx}\right)$ the equation is not a polynomial in its derivatives, so its degree is not defined.
2. The order and degree of a differential equation, when the degree is defined, are always positive integers.
Answer: True. Order and degree (when it exists) are always positive integers.
3. The general solution of a third-order differential equation contains three arbitrary constants.
Answer: True. The number of arbitrary constants equals the order of the equation.
4. Every first-order linear differential equation is a homogeneous differential equation.
Answer: False. A linear equation $\dfrac{dy}{dx} + Py = Q$ need not be homogeneous; for example $\dfrac{dy}{dx} + y = \sin x$ is linear but not homogeneous.
5. In the variable-separable method, the terms containing $y$ are kept with $dy$ and those containing $x$ are kept with $dx$.
Answer: True. That is exactly how the variables are separated before integrating.
Short Answer Questions
1. Find the order and degree of $\left(\dfrac{d^3y}{dx^3}\right)^2 – 3\dfrac{d^2y}{dx^2} + 2\left(\dfrac{dy}{dx}\right)^4 = y$.
Answer: The highest-order derivative is $\dfrac{d^3y}{dx^3}$, so the order is $3$. It is a polynomial in the derivatives and the highest power of $\dfrac{d^3y}{dx^3}$ is $2$, so the degree is $2$.
2. Find the general solution of $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$.
Answer: Separating, $\dfrac{dy}{1 + y^2} = \dfrac{dx}{1 + x^2}$. Integrating, $\tan^{-1} y = \tan^{-1} x + C$, that is $\tan^{-1} y – \tan^{-1} x = C$.
3. Find the integrating factor of $x\dfrac{dy}{dx} + 2y = x^2$.
Answer: Dividing by $x$, $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$, so $P = \dfrac{2}{x}$ and $\text{I.F.} = e^{\int 2\,dx/x} = e^{2\log x} = x^2$.
4. Form the differential equation representing the family of straight lines $y = mx$, where $m$ is an arbitrary constant.
Answer: From $y = mx$ we have $m = \dfrac{y}{x}$. Differentiating, $\dfrac{dy}{dx} = m = \dfrac{y}{x}$. Hence the required differential equation is $x\dfrac{dy}{dx} – y = 0$.
Key Terms
| Term | Meaning |
|---|---|
| Differential equation | An equation involving the derivatives of a dependent variable with respect to one or more independent variables. |
| Ordinary differential equation | A differential equation involving derivatives with respect to only one independent variable. |
| Order | The order of the highest-order derivative appearing in the differential equation. |
| Degree | The highest power (a positive integer) of the highest-order derivative, defined only when the equation is a polynomial in its derivatives. |
| General solution | A solution containing as many arbitrary constants as the order of the differential equation. |
| Particular solution | A solution free from arbitrary constants, obtained by giving the constants particular values. |
| Solution curve | The curve $y = \phi(x)$ representing a solution of the differential equation; also called an integral curve. |
| Variables separable | A method for equations that can be written as $\frac{1}{h(y)}\,dy = g(x)\,dx$, then integrated on both sides. |
| Homogeneous function of degree $n$ | A function $F$ with $F(\lambda x, \lambda y) = \lambda^n F(x, y)$ for every non-zero $\lambda$. |
| Homogeneous differential equation | An equation $\frac{dy}{dx} = F(x, y)$ where $F$ is homogeneous of degree zero; solved by putting $y = vx$. |
| Linear differential equation | An equation of the form $\frac{dy}{dx} + Py = Q$, with $P, Q$ functions of $x$ only. |
| Integrating factor (I.F.) | The factor $e^{\int P\,dx}$ that makes the left side of a linear equation an exact derivative. |