Question

Find the general solution of differential equation.

$sinx\frac{dy}{dx}+3y=cosx$

• A. $\left(\frac{1}{3}+y\right)ta{n}^3\frac{x}{2}=c+2tan\frac{x}{2}-x$
• B. $\left(\frac{1}{3}-y\right)co{t}^3\frac{x}{2}=c+2cot\frac{x}{2}-x$
• C. $\left(\frac{1}{3}-y\right)ta{n}^3\frac{x}{2}=c+2tan\frac{x}{2}+x$
• D. $\left(\frac{1}{3}+y\right)co{t}^3\frac{x}{2}=c+2cot\frac{x}{2}+x$

Answer 1

The correct option is A $\left(\frac{1}{3}+y\right)ta{n}^3\frac{x}{2}=c+2tan\frac{x}{2}-x$

$sinx\frac{dy}{dx}+3y=cosx$
$\Rightarrow \frac{dy}{dx}+\frac{3y}{\sin x}=\cot x$
which is a linear differential equation
Here, $P=3cosecx,Q=\cot x$
Integrating factor $I.F.=e^{\int Pdx}$
$=e^{\int 3cosecxdx}$
$=e^{3\log \tan \frac{x}{2}}$
$\Rightarrow I.F.={\tan }^3\frac{x}{2}$
Solution of given differential eqn is given by
$y{\tan }^3\frac{x}{2}=\int \frac{{\tan }^3\frac{x}{2}}{\tan x}dx+C$
$y{\tan }^3\frac{x}{2}=\frac{1}{2}\int {\tan }^2\frac{x}{2}(1-{\tan }^2\frac{x}{2})dx+C$
$\Rightarrow y{\tan }^3\frac{x}{2}=\frac{1}{2}\int {\tan }^2\frac{x}{2}dx-\frac{1}{2}\int {\tan }^2\frac{x}{2}{\tan }^2\frac{x}{2}dx+C$
$\Rightarrow y{\tan }^3\frac{x}{2}=\frac{1}{2}\int {\tan }^2\frac{x}{2}dx-\frac{1}{2}\int {\tan }^2\frac{x}{2}({\sec }^2\frac{x}{2}-1)dx+C$
$\Rightarrow y{\tan }^3\frac{x}{2}=\frac{1}{2}\int {\tan }^2\frac{x}{2}dx-\frac{1}{2}\int {\tan }^2\frac{x}{2}{\sec }^2\frac{x}{2}dx+\frac{1}{2}\int {\tan }^2\frac{x}{2}dx+C$
$y{\tan }^3\frac{x}{2}=\int {\tan }^2\frac{x}{2}dx-\frac{1}{2}\int {\tan }^2\frac{x}{2}{\sec }^2\frac{x}{2}dx+C$
Substitute $\tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
$\Rightarrow y{\tan }^3\frac{x}{2}=\int ({\sec }^2\frac{x}{2}-1)dx-\int t^{2}dt+C$
$\Rightarrow y{\tan }^3\frac{x}{2}=2\tan \frac{x}{2}-x-\frac{1}{3}{\tan }^3\frac{x}{2}+C$
$\Rightarrow (y+\frac{1}{3}){\tan }^3\frac{x}{2}=2\tan \frac{x}{2}-x+C$