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 differential equation $\frac{dy}{dx}+f\left(x\right)y=r\left(x\right)$ is

• A. $\frac{1}{y\left(x\right)}\int y_1\left(x\right)r\left(x\right)dx$
• B. $y_1\left(x\right)\int \frac{r\left(x\right)}{y_1\left(x\right)}dx$
• C. $fr\left(x\right)y_1\left(x\right)dx$
• D. None of these

Answer 1

Answer: (B)

$y_1\left(x\right)\int \frac{r\left(x\right)}{y_1\left(x\right)}dx$

Given the differential equation $\frac{dy}{dx}+f\left(x\right)y=0$.

Solving this equation we get, $y\left(x\right)=c{e}^{-\int f\left(x\right)dx}=y_1\left(x\right)$ [ Given].....(1).

Now, $\frac{dy}{dx}+f\left(x\right)y=r\left(x\right)$....(2)

Integrating factor (I.F.) of the above equation we get, $e^{\int f\left(x\right)dx}$.

Now multiplying the equation (2) with integrating factor and then integrating we get,

$y{e}^{\int f\left(x\right)dx}=\int e^{\int f\left(x\right)dx}r\left(x\right)dx$

or, $y=e^{-\int f\left(x\right)dx}.\int e^{\int f\left(x\right)dx}r\left(x\right)dx$

or, $y=c{e}^{-\int f\left(x\right)dx}.\int \frac{e^{\int f\left(x\right)dx}}{c}r\left(x\right)dx$

or, $y=y_1\left(x\right)\int \frac{r\left(x\right)}{y_1\left(x\right)}dx$