Question

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

• A. $\frac{1}{y(x)}\int y_1(x)dx$
• B. $y_1(x)\int \frac{r(x)}{y_1(x)}dx$
• C. $\int r(x)y_1(x)dx$
• D. $\int (r(x){)}^2{y}_1(x)dx$

Answer 1

Answer: (B)

(1) $\frac{d{y}_1}{dx}+f(x)y_1=0$
$\Rightarrow f(x)=\frac{-1}{y_1}\frac{d{y}_1}{dx}$
(2) $\frac{dy}{dx}-\frac{1}{y_1}\frac{d{y}_1}{dx}\cdot y=r(x)$
$e^{-\int \frac{1}{y_1}\frac{d_1}{dx}dx}=e^{-\int \frac{d{y}_1}{y_1}}=\frac{1}{y_1}$
$\frac{d}{dx}\left(\frac{y}{y_1}\right)=\frac{r(x)}{y_1}$
$\Rightarrow \frac{y}{y_1}=\int \frac{r(x)dx}{y_1}+c$
$y=y_1\int \frac{r(x)dx}{y_1}+c{y}_1$