Question

The solution of the differential equation $x\frac{dy}{dx}+y=x\cos \left(x\right)+\sin \left(x\right)$, given that $y=1$ when $x=\frac{\pi }{2}$ is

• A. $y=\sin \left(x\right)-\cos \left(x\right)$
• B. $y=\cos \left(x\right)$
• C. $y=\sin \left(x\right)$
• D. $y=\sin \left(x\right)+\cos \left(x\right)$

Answer 1

The correct option is C

$y=\sin \left(x\right)$

Explanation of the correct option.

Compute the required value.

Given :$x\frac{dy}{dx}+y=x\cos \left(x\right)+\sin \left(x\right)$

$\Rightarrow$ $\frac{dy}{dx}+\frac{y}{x}=\cos \left(x\right)+\sin \left(x\right)$

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

$P=\frac{1}{x}$ and $Q=\cos \left(x\right)+\sin \left(x\right)$

Now,

$\begin{array}{rcl}\mathrm{I}.\mathrm{F}.& =& {\mathrm{e}}^{\int \mathrm{P}\mathrm{dx}}\\ & =& {\mathrm{e}}^{\int \frac{1}{\mathrm{x}}\mathrm{dx}}\left[\int \frac{1}{\mathrm{x}}\mathrm{dx}=\ln \left(\mathrm{x}\right)\right]\\ & =& {\mathrm{e}}^{\ln \left(\mathrm{x}\right)}\\ & =& \mathrm{x}\end{array}$

Since the solution of L.D.E. is given by $y(\mathrm{I}.\mathrm{F}.)=\int Q\left(\mathrm{I}.\mathrm{F}.\right)dx+c$

$y.x=\int \left(x\cos \left(x\right)+\sin \left(x\right)\right)dx+c$

$\Rightarrow$$y.x=x.\sin \left(x\right)+c$

Since $y=1$when $x=\frac{\pi }{2}$

$\Rightarrow$ $\frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{2}+c$

$\Rightarrow$ $c=0$

Therefore, $y=\sin \left(x\right)$

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