Question

If $y_1\left(x\right)$ is a solution of the differential equation $\frac{dy}{dx}-f\left(x\right)y=0$, then a solution of the differential equation $\frac{dy}{dx}+f\left(x\right)y=r\left(x\right)$

• A. $\frac{1}{y_1\left(x\right)}\int r\left(x\right)y_1\left(x\right)dx$
• B. $y_1\left(x\right)\int \frac{r\left(x\right)}{y_1\left(x\right)}dx$
• C. $\int r\left(x\right)y_1\left(x\right)dx$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{1}{y_1\left(x\right)}\int r\left(x\right)y_1\left(x\right)dx$

$\frac{dy}{dx}-f\left(x\right).y=0\frac{dy}{y}=f\left(x\right)dx$
ln $y=\int f\left(x\right)dx$
$y_1\left(x\right)=e^{\int f\left(x\right)dx}$ Then for given equation I.F = $e^{\int f\left(x\right)dx}$
Hence Solution $y.y_1\left(x\right)=\int r\left(x\right).y_1\left(x\right)dx$
$y=\frac{1}{y_1\left(x\right)}\int r\left(x\right).y_1\left(x\right)dx$