NCERT Class 12 Mathematics — Differential Equations, Miscellaneous Exercise on Chapter 9. All 15 questions solved.

This exercise draws on all three solution methods of the chapter without saying which one applies where. Before starting a question, decide which of these it is:

What you see	Method
The equation factors into $g(x)h(y)$	Separate the variables (Exercise 9.3)
Right-hand side depends only on $\tfrac{y}{x}$ or $\tfrac{x}{y}$	Substitute $y = vx$ or $x = vy$ (Exercise 9.4)
$\tfrac{dy}{dx} + \mathrm{P}y = \mathrm{Q}$ with $\mathrm{P}, \mathrm{Q}$ in $x$	Integrating factor (Exercise 9.5)

Key insight. When none of the three patterns is visible, try swapping the roles of $x$ and $y$. Questions 10 and 12 look intractable as written and become routine the moment you treat $x$ as the dependent variable — question 10 is linear in $y$ once inverted, and question 12 separates cleanly.

Question 9 shows the other useful trick: substituting for a whole combination of the variables. With $x – y = t$ the equation collapses in two lines.

Question 1

For each of the differential equations given below, indicate its order and degree (if defined).

(i) $\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 – 6y = \log x$

(ii) $\left(\dfrac{dy}{dx}\right)^3 – 4\left(\dfrac{dy}{dx}\right)^2 + 7y = \sin x$

(iii) $\dfrac{d^4y}{dx^4} – \sin\left(\dfrac{d^3y}{dx^3}\right) = 0$

Solution.

(i) Highest derivative $\dfrac{d^2y}{dx^2}$, to the first power, in a polynomial equation. The $\log x$ is a function of $x$ only and does not matter. Order 2, degree 1.

(ii) Highest derivative $\dfrac{dy}{dx}$, appearing cubed. Order 1, degree 3.

(iii) Highest derivative $\dfrac{d^4y}{dx^4}$, so order $4$. But $\dfrac{d^3y}{dx^3}$ sits inside a sine, so the equation is not a polynomial in its derivatives. Order 4, degree not defined.

Question 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) $xy = ae^x + be^{-x} + x^2$  :  $x\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} – xy + x^2 – 2 = 0$

(ii) $y = e^x(a\cos x + b\sin x)$  :  $\dfrac{d^2y}{dx^2} – 2\dfrac{dy}{dx} + 2y = 0$

(iii) $y = x\sin 3x$  :  $\dfrac{d^2y}{dx^2} + 9y – 6\cos 3x = 0$

(iv) $x^2 = 2y^2\log y$  :  $\left(x^2 + y^2\right)\dfrac{dy}{dx} – xy = 0$

Solution.

(i) Differentiate $xy = ae^x + be^{-x} + x^2$ twice:

$$y + xy’ = ae^x – be^{-x} + 2x$$

$$y’ + y’ + xy” = ae^x + be^{-x} + 2 \quad\Longrightarrow\quad xy” + 2y’ = \left(ae^x + be^{-x}\right) + 2$$

From the original relation, $ae^x + be^{-x} = xy – x^2$. Substituting:

$$xy” + 2y’ = xy – x^2 + 2$$

which rearranges to the required equation.

(ii) $y’ = e^x(a\cos x + b\sin x) + e^x(-a\sin x + b\cos x) = y + e^x(b\cos x – a\sin x)$. Differentiating again:

$$y” = y’ + e^x(b\cos x – a\sin x) + e^x(-b\sin x – a\cos x) = y’ + (y’ – y) – y = 2y’ – 2y$$

using $e^x(b\cos x – a\sin x) = y’ – y$ and $e^x(a\cos x + b\sin x) = y$. Hence $y” – 2y’ + 2y = 0$.

(iii) $y’ = \sin 3x + 3x\cos 3x$, and

$$y” = 3\cos 3x + 3\cos 3x – 9x\sin 3x = 6\cos 3x – 9y$$

so $y” + 9y – 6\cos 3x = 0$.

(iv) Differentiate $x^2 = 2y^2\log y$ implicitly:

$$2x = 4yy’\log y + 2y^2 \cdot \frac{y’}{y} = 2y'(2y\log y + y)$$

From the original relation, $2y\log y = \tfrac{x^2}{y}$, so the bracket is $\tfrac{x^2}{y} + y = \tfrac{x^2+y^2}{y}$:

$$2x = 2y’\frac{x^2+y^2}{y} \quad\Longrightarrow\quad \left(x^2+y^2\right)y’ = xy$$

Question 3

Prove that $x^2 – y^2 = c\left(x^2 + y^2\right)^2$ is the general solution of the differential equation $\left(x^3 – 3xy^2\right)dx = \left(y^3 – 3x^2y\right)dy$, where $c$ is a parameter.

Solution. The equation is homogeneous of degree $3$ on both sides, so put $y = vx$:

$$\frac{dy}{dx} = \frac{x^3 – 3xy^2}{y^3 – 3x^2y} = \frac{1 – 3v^2}{v^3 – 3v}$$

$$x\frac{dv}{dx} = \frac{1 – 3v^2 – v^4 + 3v^2}{v^3 – 3v} = \frac{1 – v^4}{v^3 – 3v}$$

$$\frac{\left(v^3 – 3v\right)dv}{1 – v^4} = \frac{dx}{x}$$

The numerator is $v\left(v^2 – 3\right)dv$, and $v\,dv$ is half of $d\left(v^2\right)$, so put $t = v^2$, $dt = 2v\,dv$. The left side becomes

$$\frac{(t – 3)\,dt}{2(1 – t)(1 + t)}$$

Writing $t – 3 = \mathrm{A}(1 + t) + \mathrm{B}(1 – t)$ and putting $t = 1$ and $t = -1$ gives $\mathrm{A} = -1$ and $\mathrm{B} = -2$:

$$\frac{t – 3}{(1 – t)(1 + t)} = -\frac{1}{1 – t} – \frac{2}{1 + t}$$

Integrating both sides of the separated equation:

$$\frac12\log|1 – t| – \log(1 + t) = \log|x| + \log k$$

Multiplying by $2$ and combining the logarithms:

$$\frac{1 – t}{(1 + t)^2} = c\,x^2 \qquad \left(c = \pm k^2\right)$$

Now put back $t = \tfrac{y^2}{x^2}$. The left side is $\dfrac{\left(x^2 – y^2\right)/x^2}{\left(x^2 + y^2\right)^2/x^4} = \dfrac{x^2\left(x^2 – y^2\right)}{\left(x^2 + y^2\right)^2}$, and cancelling $x^2$ from both sides leaves

$$x^2 – y^2 = c\left(x^2 + y^2\right)^2$$

It carries one arbitrary constant, as the general solution of a first-order equation must.

Question 4

Find the general solution of the differential equation $\dfrac{dy}{dx} + \sqrt{\dfrac{1-y^2}{1-x^2}} = 0$.

Solution. The variables are already separated once the root is split:

$$\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$$

Integrating both sides:

Question 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 $(x + y + 1) = \mathrm{A}\left(1 – x – y – 2xy\right)$, where $\mathrm{A}$ is a parameter.

Solution. Separating:

$$\frac{dy}{y^2+y+1} = -\frac{dx}{x^2+x+1}$$

Each denominator completes to $\left(t + \tfrac12\right)^2 + \tfrac34$, so both integrals are arctangents with $a = \tfrac{\sqrt3}{2}$:

$$\frac{2}{\sqrt3}\tan^{-1}\frac{2y+1}{\sqrt3} + \frac{2}{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3} = \mathrm{C}$$

Combining the two arctangents with the addition formula $\tan^{-1}p + \tan^{-1}q = \tan^{-1}\tfrac{p+q}{1-pq}$ and simplifying:

$$\frac{\frac{2x+1}{\sqrt3} + \frac{2y+1}{\sqrt3}}{1 – \frac{(2x+1)(2y+1)}{3}} = \frac{2\sqrt3(x+y+1)}{3 – (2x+1)(2y+1)} = \frac{2\sqrt3(x+y+1)}{2 – 2x – 2y – 4xy}$$

which is a constant multiple of $\dfrac{x+y+1}{1 – x – y – 2xy}$. Setting that ratio equal to a parameter $\mathrm{A}$ gives the stated form.

Question 6

Find the equation of the curve passing through the point $\left(0, \dfrac{\pi}{4}\right)$ whose differential equation is $\sin x\cos y\,dx + \cos x\sin y\,dy = 0$.

Solution. Divide by $\cos x\cos y$ to separate:

$$\tan x\,dx + \tan y\,dy = 0$$

$$\log|\sec x| + \log|\sec y| = \log \mathrm{C} \quad\Longrightarrow\quad \sec x\sec y = \mathrm{C}$$

At $\left(0, \tfrac{\pi}{4}\right)$: $\sec 0 \cdot \sec\tfrac{\pi}{4} = \sqrt2 = \mathrm{C}$. So $\sec x\sec y = \sqrt2$, that is $\cos y = \dfrac{\sec x}{\sqrt2}$.

Question 7

Find the particular solution of the differential equation $\left(1 + e^{2x}\right)dy + \left(1 + y^2\right)e^x dx = 0$, given that $y = 1$ when $x = 0$.

Solution. Separate:

$$\frac{dy}{1+y^2} = -\frac{e^x\,dx}{1+e^{2x}}$$

On the right, put $t = e^x$, so $dt = e^x dx$ and the integral is $\int\tfrac{dt}{1+t^2} = \tan^{-1}\left(e^x\right)$:

$$\tan^{-1}y = -\tan^{-1}\left(e^x\right) + \mathrm{C}$$

At $x = 0$, $y = 1$: $\tfrac{\pi}{4} = -\tfrac{\pi}{4} + \mathrm{C}$, so $\mathrm{C} = \tfrac{\pi}{2}$.

Question 8

Solve the differential equation $y\,e^{x/y}dx = \left(x\,e^{x/y} + y^2\right)dy$  $(y \ne 0)$.

Solution. The ratio appearing is $\tfrac{x}{y}$, so substitute $x = vy$ with $\dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$:

$$v + y\frac{dv}{dy} = \frac{vy\,e^v + y^2}{y\,e^v} = v + \frac{y}{e^v}$$

The $v$ cancels, leaving

$$y\frac{dv}{dy} = \frac{y}{e^v} \quad\Longrightarrow\quad e^v\,dv = dy$$

$$e^v = y + \mathrm{C}$$

Question 9

Find a particular solution of the differential equation $(x-y)(dx + dy) = dx – dy$, given that $y = -1$ when $x = 0$.  [Hint: put $x – y = t$]

Solution. Expanding and collecting:

$$(x-y)\,dx + (x-y)\,dy = dx – dy \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1 – (x-y)}{1 + (x-y)}$$

Following the hint, put $t = x – y$, so $\dfrac{dt}{dx} = 1 – \dfrac{dy}{dx}$:

$$\frac{dt}{dx} = 1 – \frac{1-t}{1+t} = \frac{1 + t – 1 + t}{1+t} = \frac{2t}{1+t}$$

$$\frac{(1+t)\,dt}{2t} = dx \quad\Longrightarrow\quad \frac12\left(\log|t| + t\right) = x + c$$

Substituting $t = x – y$ and doubling:

$$\log|x – y| + (x – y) = 2x + \mathrm{C} \quad\Longrightarrow\quad \log|x-y| = x + y + \mathrm{C}$$

At $(0, -1)$: $\log 1 = 0 = 0 – 1 + \mathrm{C}$, so $\mathrm{C} = 1$.

Question 10

Solve the differential equation $\left[\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} – \dfrac{y}{\sqrt{x}}\right]\dfrac{dx}{dy} = 1$  $(x \ne 0)$.

Solution. Invert to get a linear equation in $y$:

$$\frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} – \frac{y}{\sqrt{x}} \quad\Longrightarrow\quad \frac{dy}{dx} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$$

$$\int\frac{dx}{\sqrt{x}} = 2\sqrt{x} \quad\Longrightarrow\quad \mathrm{I.F.} = e^{2\sqrt{x}}$$

$$y\,e^{2\sqrt{x}} = \int e^{2\sqrt{x}} \cdot \frac{e^{-2\sqrt{x}}}{\sqrt{x}}dx = \int\frac{dx}{\sqrt{x}} = 2\sqrt{x} + \mathrm{C}$$

The two exponentials cancelling is what makes this question short.

Question 11

Find a particular solution of the differential equation $\dfrac{dy}{dx} + y\cot x = 4x\operatorname{cosec}x$  $(x \ne 0)$, given that $y = 0$ when $x = \dfrac{\pi}{2}$.

Solution. Linear with $\mathrm{P} = \cot x$, so $\mathrm{I.F.} = e^{\log|\sin x|} = \sin x$:

$$y\sin x = \int 4x\operatorname{cosec}x \cdot \sin x\,dx = \int 4x\,dx = 2x^2 + \mathrm{C}$$

At $x = \tfrac{\pi}{2}$, $y = 0$: $0 = 2\left(\tfrac{\pi}{2}\right)^2 + \mathrm{C}$, so $\mathrm{C} = -\tfrac{\pi^2}{2}$.

Question 12

Find a particular solution of the differential equation $(x+1)\dfrac{dy}{dx} = 2e^{-y} – 1$, given that $y = 0$ when $x = 0$.

Solution. Separate:

$$\frac{dy}{2e^{-y} – 1} = \frac{dx}{x+1}$$

Multiply numerator and denominator on the left by $e^y$:

$$\frac{e^y\,dy}{2 – e^y} = \frac{dx}{x+1}$$

The left numerator is minus the derivative of the denominator:

$$-\log\left|2 – e^y\right| = \log|x+1| + \log c$$

$$\frac{1}{2 – e^y} = c(x+1)$$

At $(0, 0)$: $\tfrac{1}{2-1} = 1 = c$, so $2 – e^y = \tfrac{1}{x+1}$ and

$$e^y = 2 – \frac{1}{x+1} = \frac{2x+1}{x+1}$$

Question 13

The general solution of the differential equation $\dfrac{y\,dx – x\,dy}{y} = 0$ is

Solution. Since the fraction is zero, its numerator is: $y\,dx – x\,dy = 0$, so

$$\frac{dy}{y} = \frac{dx}{x} \quad\Longrightarrow\quad \log|y| = \log|x| + \log \mathrm{C}$$

Question 14

The general solution of a differential equation of the type $\dfrac{dx}{dy} + \mathrm{P}_1x = \mathrm{Q}_1$ is

Solution. Two things must match the equation’s form. The dependent variable is $x$, so it is $x$ that multiplies the integrating factor — this rules out (A) and (B). And $\mathrm{P}_1$ and $\mathrm{Q}_1$ are functions of $y$, so every integration is with respect to $y$ — this rules out (D).

Question 15

The general solution of the differential equation $e^x\,dy + \left(y\,e^x + 2x\right)dx = 0$ is

Solution. Notice that $e^x\,dy + y\,e^x\,dx$ is exactly $d\!\left(y\,e^x\right)$, by the product rule. So the equation reads

$$d\!\left(y\,e^x\right) + 2x\,dx = 0$$

Integrating directly:

$$y\,e^x + x^2 = \mathrm{C}$$

The same answer follows from the standard linear method: dividing by $e^x$ gives $\tfrac{dy}{dx} + y = -2xe^{-x}$ with $\mathrm{I.F.} = e^x$ — but spotting the exact differential is a line and a half shorter.

Common mistakes

• Committing to a method before classifying the equation. Questions 8, 10 and 12 all look like one type and are another. Spend ten seconds on the table above before writing anything.
• Not inverting when the equation is linear in $x$ rather than $y$. Question 10 is unsolvable as written and routine once flipped.
• Missing an exact differential. Question 15 is $d(ye^x) + 2x\,dx = 0$; seeing it saves the whole integrating-factor calculation.
• Substituting for a single variable when a combination is what repeats. In question 9, $x – y$ appears three times; substituting $t = x – y$ is the hint the book gives, and it is the only practical route.
• Applying the initial condition to an unsimplified form. In question 12 the condition is far easier to apply after solving for $e^y$.
• Forgetting that $\mathrm{P}$ and $\mathrm{Q}$ must be functions of the independent variable. Question 14 is entirely about which variable is which.

Practise next

• Exercise 9.5 — the integrating factor used in questions 10, 11 and 15.
• Exercise 9.4 — the homogeneous substitution used in questions 3 and 8.