CBSE Board Examination 2024 · Section C · 3 marks

Q29: Particular solution of $x^2\frac{dy}{dx} - xy = x^2\cos^2\left(\frac{y}{2x}\right)$ ⁠, $y(1) = \frac{\pi}{2}$

Board 2024Ch 9 · Differential Equations

The question

Find the particular solution of the differential equation given by $x^2\dfrac{dy}{dx} - xy = x^2\cos^2\left(\dfrac{y}{2x}\right)$ ⁠, given that when $x = 1$ ⁠, $y = \dfrac{\pi}{2}$ ⁠.

Exam answer

Dividing by $x^2$ ⁠,

\frac{dy}{dx} = \frac{y}{x} + \cos^2\left(\frac{y}{2x}\right)

Why?

First get $\frac{dy}{dx}$ on its own. Dividing every term by $x^2$ ⁠: $\frac{xy}{x^2} = \frac{y}{x}$ ⁠, and the last term loses its $x^2$ ⁠.

\begin{aligned} \frac{dy}{dx} - \frac{xy}{x^2} &= \cos^2\frac{y}{2x} \\ \frac{dy}{dx} - \frac{y}{x} &= \cos^2\frac{y}{2x} \\ \frac{dy}{dx} &= \frac{y}{x} + \cos^2\frac{y}{2x} \end{aligned}

(The last move adds $\frac{y}{x}$ to both sides.)

This is a homogeneous equation. Put $y = vx$ ⁠, so that $\frac{dy}{dx} = v + x\frac{dv}{dx}$ ⁠.

Why?

The right side depends only on the ratio $\frac{y}{x}$ ⁠, so the equation is homogeneous. The standard method is $y = vx$ ⁠, where $v$ is a function of $x$ ⁠.

The product rule gives $\frac{dy}{dx} = v\cdot 1 + x\frac{dv}{dx}$ ⁠. Also $\frac{y}{x} = v$ and $\frac{y}{2x} = \frac{v}{2}$ ⁠.

\begin{aligned} v + x\frac{dv}{dx} &= v + \cos^2\frac{v}{2} \\ \Rightarrow x\frac{dv}{dx} &= \cos^2\frac{v}{2} \\ \Rightarrow \sec^2\frac{v}{2}\,dv &= \frac{dx}{x} \end{aligned}

Why?

The $v$ on each side cancels. Then separate the variables: all the $v$ terms with $dv$ ⁠, all the $x$ terms with $dx$ ⁠. Divide by $\cos^2\frac{v}{2}$ and by $x$ ⁠, using $\frac{1}{\cos^2\alpha} = \sec^2\alpha$ ⁠.

\begin{aligned} x\frac{dv}{dx} &= \cos^2\frac{v}{2} \quad (\text{subtract } v) \\ \frac{dv}{\cos^2\frac{v}{2}} &= \frac{dx}{x} \\ \sec^2\frac{v}{2}\,dv &= \frac{dx}{x} \end{aligned}

Integrating both sides,

\begin{aligned} \int\sec^2\frac{v}{2}\,dv &= \int\frac1x\,dx \\ \Rightarrow 2\tan\frac{v}{2} &= \log|x| + C \\ \Rightarrow 2\tan\left(\frac{y}{2x}\right) &= \log|x| + C \end{aligned}

Why?

For a constant $k \ne 0$ ⁠,

\int\sec^2 kv\,dv = \frac1k\tan kv

Here $k = \frac12$ ⁠, so $\frac1k = 2$ ⁠. Also $\int\frac1x\,dx = \log|x|$ ⁠, and one constant C covers both sides.

Then put back $v = \frac{y}{x}$ ⁠, so $\frac{v}{2} = \frac{y}{2x}$ ⁠.

When $x = 1$ ⁠, $y = \frac{\pi}{2}$ ⁠: $2\tan\frac{\pi}{4} = \log 1 + C$ ⁠, so $C = 2$ ⁠.

Why?

The given values fix C, which turns the general solution into the particular one. At $x = 1$ ⁠, $\frac{y}{2x} = \frac{\pi}{4}$ ⁠. Then $\tan\frac{\pi}{4} = 1$ and $\log 1 = 0$ ⁠, so $2 = 0 + C$ ⁠.

\begin{aligned} \frac{y}{2x} &= \frac{\pi/2}{2(1)} = \frac{\pi}{2} \times \frac12 = \frac{\pi}{4} \\[4pt] 2\tan\frac{\pi}{4} &= \log|1| + C \\ 2(1) &= 0 + C \\ C &= 2 \end{aligned}

$\therefore\ 2\tan\left(\dfrac{y}{2x}\right) = \log|x| + 2$ is the required particular solution.

Where this comes from

Chapter 9, Differential Equations: Homogeneous equation: particular solution.

Question type: Homogeneous equations + A particular solution (combines 2 types)

How to spot it: click to see

Type 6. Homogeneous equations: put $y = vx$

Spot it: every term has the same degree, like $x^2$ ⁠, $xy$ and $y^2$ ⁠, or you see $\frac{y}{x}$ inside sin, cos or log. You cannot separate $x$ and $y$ ⁠.

Here: Dividing by $x^2$ gives $\frac{dy}{dx} = \frac{y}{x} + \cos^2\frac{y}{2x}$ ⁠. Put $y = vx$ ⁠: the $v$ cancels, leaving $\sec^2\frac{v}{2}\,dv = \frac{dx}{x}$ ⁠.

Type 4. A particular solution: solve, then find $C$

Spot it: a condition such as " $y = 1$ when $x = 0$ ", or "the curve passes through a point".

Here: Then put $x = 1$ ⁠, $y = \frac{\pi}{2}$ ⁠: $2\tan\frac{\pi}{4} = \log 1 + C$ ⁠, so $C = 2$ ⁠.

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Differential Equations in past papers: question 8 of 31 · see them all

Q29: Particular solution of x2dydx−xy=x2cos⁡2(y2x)x^2\frac{dy}{dx} - xy = x^2\cos^2\left(\frac{y}{2x}\right)x2dxdy​−xy=x2cos2(2xy​)⁠, y(1)=π2y(1) = \frac{\pi}{2}y(1)=2π​

Q29: Particular solution of $x^2\frac{dy}{dx} - xy = x^2\cos^2\left(\frac{y}{2x}\right)$ ⁠, $y(1) = \frac{\pi}{2}$