Question

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

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

Answer 1

The correct option is D

$y\cos \left(x\right)=c+\frac{1}{2}\cos \left(2x\right)$

Explanation of the correct option.

Compute the required value.

Given : $\frac{dy}{dx}=y\tan \left(x\right)-2\sin \left(x\right)$

$\Rightarrow$$\frac{dy}{dx}-y\tan \left(x\right)=-2\sin \left(x\right)$……………$\left(1\right)$

Compare the equation with L.D.E. $\frac{dy}{dx}+Py=Q$,

$P=-\tan \left(x\right)$ and $Q=-2\sin \left(x\right)$

$\begin{array}{rcl}\mathrm{I}.\mathrm{F}.& =& {\mathrm{e}}^{\int \mathrm{P}\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\int \tan \left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\log \left(\left|\sec \left(\mathrm{x}\right)\right|\right)}\\ & =& \cos \left(\mathrm{x}\right)\end{array}$

Since the solution of L.D.E. is given by $y(\mathrm{I}.\mathrm{F}.)=\int Qdx$.

$$\begin{gathered} y.\cos \left(x\right)=-\int \sin \left(2x\right)dx+c \\ \Rightarrow y.\cos \left(x\right)=\frac{\cos \left(2x\right)}{2}+c \end{gathered}$$ $\left[\int \sin \left(x\right)=-\cos \left(x\right)\right]$

Hence, option $\left(D\right)$ is the correct option.