Question

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

• A. $\frac{1}{y(x)} \int y_1(x) d x$
• B. $y_1(x) \int \frac{r(x)}{y_1(x)} d x$
• C. $\int r(x) y_1(x) d x$
• D. $\int(r(x))^2 y_1(x) d x$

Answer 1

Answer: (B)

(1) $\frac{d y_1}{d x}+f(x) y_1=0 $
$ \Rightarrow f(x)=\frac{-1}{y_1} \frac{d y_1}{d x}$
(2) $ \frac{d y}{d x}-\frac{1}{y_1} \frac{d y_1}{d x} \cdot y=r(x)$
$ e^{-\int \frac{1}{y_1} \frac{d_1}{d x} d x}=e^{-\int \frac{d y_1}{y_1}}=\frac{1}{y_1} $
$ \frac{d}{d x}\left(\frac{y}{y_1}\right)=\frac{r(x)}{y_1} $
$ \Rightarrow \frac{y}{y_1}=\int \frac{r(x) d x}{y_1}+c$
$ y=y_1 \int \frac{r(x) d x}{y_1}+c y_1$