Question

The solution of the differential equation $\frac{dy}{dx}+\frac{\tan \left(y\right)}{x}=\frac{\tan \left(y\right)\sin \left(y\right)}{x^2}$ is

• A. $2x=\sin \left(y\right)\left(1+c{x}^2\right)$
• B. $\frac{y}{\sin \left(x\right)}+\log \left(x\right)=c$
• C. $\log \left(x\right)+x=c$
• D. $\log \left(x\right)+y=c$

Answer 1

The correct option is A

$2x=\sin \left(y\right)\left(1+c{x}^2\right)$

Explanation for correct option:

Given differential equation is $\frac{dy}{dx}+\frac{\tan \left(y\right)}{x}=\frac{\tan \left(y\right)\sin \left(y\right)}{x^2}$

Therefore,

$\begin{array}{rcl}\frac{dy}{dx}+\frac{\tan \left(y\right)}{x}& =& \frac{\tan \left(y\right)\sin \left(y\right)}{x^2}\\ & \Rightarrow & \frac{1}{\tan \left(y\right)\sin \left(y\right)}\frac{dy}{dx}+\frac{\tan \left(y\right)}{x\tan \left(y\right)\sin \left(y\right)}=\frac{1}{x^2}\\ & \Rightarrow & \cot \left(y\right)cosec\left(y\right)\frac{dy}{dx}+\frac{cosec\left(y\right)}{x}=\frac{1}{x^2}\end{array}$

Let $cosec\left(y\right)=t$.

On differentiating both sides we get $-cosec\left(y\right)\cot \left(y\right)dy=dt$

Now substituting we get

$\Rightarrow \frac{dt}{dx}-\frac{t}{x}=-\frac{1}{x^2}$

We know that for the differential equation $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ the integrating factor is $f\left(x\right)=e^{\int P\left(x\right)dx}$.

Then the solution of the Differential equation is $\int d\left(yf\left(x\right)\right)=\int Q\left(x\right)f\left(x\right)dx+c$

Now comparing with the standard form we have $P\left(x\right)=-\frac{1}{x},Q\left(x\right)=-\frac{1}{x^2}$

Therefore Integrating factor

$$\begin{gathered} f\left(x\right)=e^{\int -\frac{1}{x}dx} \\ =e^{-\log \left(x\right)} \\ =\frac{1}{x} \end{gathered}$$

Therefore the solution is

$\begin{array}{rcl}\int dt\left(\frac{1}{x}\right)& =& -\int \frac{1}{x^2}·\frac{1}{x}dx+c\\ & \Rightarrow & \frac{t}{x}=-\int \frac{1}{x^3}dx+c\\ & \Rightarrow & \frac{t}{x}=\frac{1}{2x^2}+c\\ & \Rightarrow & cosec\left(y\right)=\frac{1}{2x}+c{x}^2\left[\because t=cosec\left(y\right)\right]\\ & \Rightarrow & 2x=\sin \left(y\right)\left(1+c{x}^2\right)\left[\because \frac{1}{\sin \left(y\right)}=cosec\left(y\right)\right]\end{array}$

Hence, option $\left(A\right)$ i.e. $2x=\sin \left(y\right)\left(1+c{x}^2\right)$ is correct