Question

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

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

Answer 1

The correct option is A

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

Explanation of the correct option.

Compute the required value.

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

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

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

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

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

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

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