CBSE Board Examination 2022 (Term 2) · Section B · 3 marks

Q10: Particular solution of $x\frac{dy}{dx} + x\cos^2\left(\frac{y}{x}\right) = y$ with $y = \frac{\pi}{4}$ at $x = 1$

Board 2022 T2Ch 9 · Differential Equations

The question

Find the particular solution of the differential equation $x\dfrac{dy}{dx} + x\cos^2\left(\dfrac{y}{x}\right) = y$ ⁠; given that when $x = 1$ ⁠, $y = \dfrac{\pi}{4}$ ⁠.

Exam answer

The given differential equation can be written as

\frac{dy}{dx} = \frac{y}{x} - \cos^2\frac{y}{x} \quad\ldots(\text{i})

Why?

Divide every term by $x$ and move $\cos^2\frac{y}{x}$ to the right, so that $\frac{dy}{dx}$ stands alone.

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

It is a homogeneous differential equation.

Why?

The right side depends only on the ratio $\frac{y}{x}$ ⁠. Replacing $x, y$ by $kx, ky$ leaves it unchanged, so it is homogeneous of degree 0.

For such equations the standard substitution is $y = vx$ ⁠, with $v$ a function of $x$ ⁠.

Let $y = vx$ ⁠. $\therefore\ \frac{dy}{dx} = v + x\frac{dv}{dx}$

Why?

Product rule on $v \cdot x$ ⁠: the derivative of $x$ is 1, which gives $v$ ⁠, plus $x$ times the derivative of $v$ ⁠.

(i) becomes

v + x\frac{dv}{dx} = v - \cos^2v

\Rightarrow x\frac{dv}{dx} = -\cos^2v

\Rightarrow \sec^2v\,dv = -\frac{dx}{x}

Why?

Since $\frac{y}{x} = v$ ⁠, the right side of (i) is $v - \cos^2v$ ⁠. The $v$ on each side cancels.

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

Separate the variables: divide by $\cos^2v$ and by $x$ ⁠, and multiply by $dx$ ⁠. Use $\frac{1}{\cos^2v} = \sec^2v$ ⁠.

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

Integrating both sides,

\int \sec^2v\,dv = -\int \frac{dx}{x}

\tan v = -\log|x| + C

Why?

Standard integrals:

\begin{aligned} \int \sec^2v\,dv &= \tan v \\ \int \frac{dx}{x} &= \log|x| \end{aligned}

One arbitrary constant $C$ is enough for both sides.

\Rightarrow \tan\frac{y}{x} = -\log|x| + C

Why?

Put back $v = \frac{y}{x}$ ⁠, since we set $y = vx$ ⁠. So $\tan v$ becomes $\tan\frac{y}{x}$ ⁠.

When $x = 1$ ⁠, $y = \frac{\pi}{4}$ ⁠:

\tan\frac{\pi}{4} = -\log 1 + C \Rightarrow C = 1

Why?

At $x = 1$ ⁠, $\frac{y}{x} = \frac{\pi}{4}$ ⁠. Then $\tan\frac{\pi}{4} = 1$ and $\log 1 = 0$ ⁠, so $1 = 0 + C$ ⁠.

\begin{aligned} \tan\frac{\pi/4}{1} &= -\log|1| + C \\ \tan\frac{\pi}{4} &= -\log 1 + C \\ 1 &= -0 + C \\ C &= 1 \end{aligned}

$\therefore$ The particular solution is $\tan\dfrac{y}{x} = -\log|x| + 1$ ⁠.

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 $x$ gives $\frac{dy}{dx} = \frac{y}{x} - \cos^2\frac{y}{x}$ ⁠, all in $\frac{y}{x}$ ⁠. Put $y = vx$ ⁠: the $v$ terms cancel, leaving $\sec^2v\,dv = -\frac{dx}{x}$ ⁠.

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: The condition $y = \frac{\pi}{4}$ at $x = 1$ then fixes the constant: $\tan\frac{\pi}{4} = -\log 1 + C$ ⁠, so $C = 1$ ⁠.

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Differential Equations in past papers: question 13 of 31 · see them all

Q10: Particular solution of xdydx+xcos⁡2(yx)=yx\frac{dy}{dx} + x\cos^2\left(\frac{y}{x}\right) = yxdxdy​+xcos2(xy​)=y with y=π4y = \frac{\pi}{4}y=4π​ at x=1x = 1x=1

Q10: Particular solution of $x\frac{dy}{dx} + x\cos^2\left(\frac{y}{x}\right) = y$ with $y = \frac{\pi}{4}$ at $x = 1$