CBSE Sample Paper 2021-22 (Term 2) · Section B · 3 marks

Q8: General solution of $x\frac{dy}{dx} = y - x\sin\left(\frac{y}{x}\right)$ ⁠, or a particular solution of $\frac{dy}{dx} + y\cot x = \frac{2}{1 + \sin x}$

Sample 2021-22 T2Ch 9 · Differential Equations

The question

(a) Find the general solution of the following differential equation:

x\frac{dy}{dx} = y - x\sin\left(\frac{y}{x}\right)

(b) Find the particular solution of the following differential equation, given that $y = 0$ when $x = \dfrac{\pi}{4}$ ⁠:

\frac{dy}{dx} + y\cot x = \frac{2}{1 + \sin x}

Exam answer

(a)

The given differential equation can be written as

\frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right)

It is a homogeneous differential equation.

Why?

Dividing both sides by $x$ leaves a right side that depends only on $\frac{y}{x}$ ⁠.

\begin{aligned} x\frac{dy}{dx} &= y - x\sin\frac{y}{x} \\ \frac{dy}{dx} &= \frac{y}{x} - \frac{x}{x}\sin\frac{y}{x} \\ &= \frac{y}{x} - \sin\frac{y}{x} \end{aligned}

An equation of the form

\frac{dy}{dx} = g\left(\frac{y}{x}\right)

is homogeneous; here $g(v) = v - \sin v$ ⁠. The standard method is the substitution $y = vx$ ⁠.

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

Why?

Here $v$ is a function of $x$ ⁠, so differentiate $vx$ by the product rule:

\frac{d}{dx}(vx) = v\cdot 1 + x\frac{dv}{dx}

The equation becomes

v + x\frac{dv}{dx} = v - \sin v

\Rightarrow x\frac{dv}{dx} = -\sin v

\Rightarrow \frac{dv}{\sin v} = -\frac{dx}{x}

Why?

Since $\frac{y}{x} = v$ ⁠, the right side is $v - \sin v$ ⁠.

The $v$ on both sides cancels. Then separate the variables: divide by $x\sin v$ ⁠, so that all the $v$ terms sit with $dv$ and all the $x$ terms with $dx$ ⁠.

\begin{aligned} x\frac{dv}{dx} &= v - \sin v - v \\ &= -\sin v \\[4pt] \frac{dv}{\sin v} &= -\frac{dx}{x} \\ \operatorname{cosec}v\,dv &= -\frac{dx}{x} \end{aligned}

The last line uses $\frac{1}{\sin v} = \operatorname{cosec}v$ ⁠.

\Rightarrow \operatorname{cosec}v\,dv = -\frac{dx}{x}

Integrating both sides,

\begin{aligned} &\log|\operatorname{cosec}v - \cot v| \\ &= -\log|x| + \log K,\ K > 0 \end{aligned}

Why?

Standard integrals:

\begin{aligned} &\int \operatorname{cosec}v\,dv \\ &= \log|\operatorname{cosec}v - \cot v| \end{aligned}

\int \frac{dx}{x} = \log|x|

The constant is written as $\log K$ with $K > 0$ ⁠, so that all the terms are logs and can be combined.

\begin{aligned} &\Rightarrow \log|(\operatorname{cosec}v - \cot v)x| \\ &\qquad = \log K \end{aligned}

\Rightarrow (\operatorname{cosec}v - \cot v)x = \pm K

Why?

Move $-\log|x|$ to the left and use $\log a + \log b = \log ab$ ⁠. Equal logs give equal numbers, so $|(\operatorname{cosec}v - \cot v)x| = K$ ⁠. Removing the modulus gives $\pm K$ ⁠.

\begin{aligned} &\log|\operatorname{cosec}v - \cot v| + \log|x| \\ &= \log K \\[4pt] &\log|(\operatorname{cosec}v - \cot v)x| \\ &= \log K \\[4pt] &|(\operatorname{cosec}v - \cot v)x| = K \end{aligned}

Finally put back $v = \frac{y}{x}$ and call the constant $\pm K$ simply $C$ ⁠.

$\therefore$ The general solution is $\left(\operatorname{cosec}\frac{y}{x} - \cot\frac{y}{x}\right)x = C$ ⁠, where $C = \pm K$ ⁠.

(b)

This is a linear differential equation $\frac{dy}{dx} + Py = Q$ with $P = \cot x$ and $Q = \frac{2}{1 + \sin x}$ ⁠.

\begin{aligned} \text{I.F.} &= e^{\int \cot x\,dx} \\ &= e^{\log\sin x} = \sin x \end{aligned}

Why?

The equation is linear: $y$ and $\frac{dy}{dx}$ appear only to the first power. For such an equation the integrating factor is

\text{I.F.} = e^{\int P\,dx}

Here $\int \cot x\,dx = \log|\sin x|$ ⁠. Near $x = \frac{\pi}{4}$ ⁠, $\sin x > 0$ ⁠, so the modulus can be dropped, and $e^{\log a} = a$ ⁠.

The general solution is given by

Why?

For $\frac{dy}{dx} + Py = Q$ the solution is

y \times \text{I.F.} = \int Q \times \text{I.F.}\,dx

plus a constant. Here $Q \times \text{I.F.} = \frac{2\sin x}{1 + \sin x}$ ⁠.

\begin{aligned} &y\sin x = \int 2\,\frac{\sin x}{1 + \sin x}\,dx \\ &= 2\int \frac{\sin x + 1 - 1}{1 + \sin x}\,dx \\ &= 2\int \left[1 - \frac{1}{1 + \sin x}\right]dx \\ &= 2\int \left[1 - \frac{1}{1 + \cos\left(\frac{\pi}{2} - x\right)}\right]dx \end{aligned}

Why?

Adding and subtracting 1 in the numerator splits the fraction:

\frac{\sin x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x}

Step by step:

\begin{aligned} &\frac{\sin x + 1 - 1}{1 + \sin x} \\ &= \frac{1 + \sin x}{1 + \sin x} - \frac{1}{1 + \sin x} \\ &= 1 - \frac{1}{1 + \sin x} \end{aligned}

Then $\sin x = \cos\left(\frac{\pi}{2} - x\right)$ ⁠, which prepares for the half-angle identity.

Using $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$ ⁠,

Why?

Take $\theta = \frac{\pi}{2} - x$ ⁠. Half of it is $\frac{\pi}{4} - \frac{x}{2}$ ⁠. Then $\frac{1}{\cos^2u} = \sec^2u$ ⁠.

\begin{aligned} \frac{\theta}{2} &= \frac12\left(\frac{\pi}{2} - x\right) \\ &= \frac{\pi}{4} - \frac{x}{2} \\[4pt] &1 + \cos\left(\frac{\pi}{2} - x\right) \\ &= 2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \end{aligned}

\begin{aligned} &\frac{1}{2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)} \\ &= \frac12\sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \end{aligned}

\begin{aligned} &y\sin x \\ &= 2\int \left[1 - \frac{1}{2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)}\right]dx \\ &= 2\int \left[1 - \frac12\sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right)\right]dx \\ &= 2\left[x + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right] + c \end{aligned}

Why?

Standard integral:

\begin{aligned} &\int \sec^2(ax + b)\,dx \\ &= \frac{1}{a}\tan(ax + b) + C \end{aligned}

Here $a = -\frac12$ ⁠, so $\frac1a = -2$ ⁠:

\begin{aligned} &\int \frac12\sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right)dx \\ &= -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \end{aligned}

Also $\int 1\,dx = x$ ⁠. So the bracket becomes

\begin{aligned} &x - \left(-\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) \\ &= x + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \end{aligned}

Subtracting this gives $+\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$ inside the bracket. Watch out: forgetting the factor $\frac1a = -2$ is the usual slip here.

Given $y = 0$ when $x = \frac{\pi}{4}$ ⁠:

0 = 2\left[\frac{\pi}{4} + \tan\frac{\pi}{8}\right] + c

\Rightarrow c = -\frac{\pi}{2} - 2\tan\frac{\pi}{8}

Why?

At $x = \frac{\pi}{4}$ ⁠: $\frac{\pi}{4} - \frac{\pi}{8} = \frac{\pi}{8}$ ⁠, and the left side is $y\sin x = 0$ ⁠.

\begin{aligned} \frac{x}{2} &= \frac{\pi/4}{2} = \frac{\pi}{8} \\ \frac{\pi}{4} - \frac{\pi}{8} &= \frac{2\pi}{8} - \frac{\pi}{8} = \frac{\pi}{8} \end{aligned}

Now solve for $c$ ⁠:

\begin{aligned} 0 &= 2\left[\frac{\pi}{4} + \tan\frac{\pi}{8}\right] + c \\ 0 &= \frac{\pi}{2} + 2\tan\frac{\pi}{8} + c \\ c &= -\frac{\pi}{2} - 2\tan\frac{\pi}{8} \end{aligned}

Then divide $y\sin x = \ldots$ by $\sin x$ ⁠, that is, multiply by $\operatorname{cosec}x$ ⁠, to get $y$ on its own.

\begin{aligned} &y\sin x \\ &= 2\left[x + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right] + c \\[4pt] &y = \operatorname{cosec}x\,\bigg[2\Big\{x \\ &\qquad + \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\Big\} + c\bigg] \end{aligned}

and $c = -\frac{\pi}{2} - 2\tan\frac{\pi}{8} = -\left(\frac{\pi}{2} + 2\tan\frac{\pi}{8}\right)$ ⁠.

Hence, the particular solution is

$y = \operatorname{cosec}x\,\Big[2\Big\{x + \tan\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\Big\}$ $- \left(\dfrac{\pi}{2} + 2\tan\dfrac{\pi}{8}\right)\Big]$

Where this comes from

Chapter 9, Differential Equations: Homogeneous and linear differential equations.

Question type: (a) Homogeneous equations; (b) Linear equations + A particular solution (combined)

How to spot it: click to see

(a) 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 $\frac{dy}{dx} = \frac{y}{x} - \sin\frac{y}{x}$ ⁠. Put $y = vx$ ⁠: the $v$ cancels, leaving $\operatorname{cosec}v\,dv = -\frac{dx}{x}$ ⁠.

(b) Type 7. Linear equations: the integrating factor

Spot it: $\frac{dy}{dx} + Py = Q$ ⁠, with $P$ and $Q$ in $x$ only, and $y$ only to power 1.

Here: Here $P = \cot x$ and $Q = \frac{2}{1 + \sin x}$ ⁠, so $\text{I.F.} = e^{\log\sin x} = \sin x$ ⁠, and $y\sin x = \int\frac{2\sin x}{1 + \sin x}\,dx$ ⁠.

(b) 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 $y = 0$ at $x = \frac{\pi}{4}$ gives $c = -\frac{\pi}{2} - 2\tan\frac{\pi}{8}$ ⁠; put it back for the particular solution.

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Differential Equations in past papers: question 28 of 31 · see them all

Q8: General solution of xdydx=y−xsin⁡(yx)x\frac{dy}{dx} = y - x\sin\left(\frac{y}{x}\right)xdxdy​=y−xsin(xy​)⁠, or a particular solution of dydx+ycot⁡x=21+sin⁡x\frac{dy}{dx} + y\cot x = \frac{2}{1 + \sin x}dxdy​+ycotx=1+sinx2​

Q8: General solution of $x\frac{dy}{dx} = y - x\sin\left(\frac{y}{x}\right)$ ⁠, or a particular solution of $\frac{dy}{dx} + y\cot x = \frac{2}{1 + \sin x}$