Question

If $y_1\&y_2$ be solutions of the differential equation $\frac{dy}{dx}+Py+Q$, where $P$ & $Q$ are functions of $x$ alone and $y_2=y_{1}z$, then $z=1-a{e}^{-\int \frac{Q}{y_1}dx}$, '$a$' being an arbitrary constant. If you think this is true write 1 otherwise write 0.

Answer 1

$\frac{d{y}_1}{dx}+P{y}_1=Q,\frac{d{y}_2}{dx}+P{y}_2=Q$
Substitue $y_2=y_{1}z$
$\Rightarrow \frac{d{y}_2}{dx}=y_1\frac{dz}{dx}+z\frac{d{y}_1}{dx}$
$\Rightarrow y_1\frac{dz}{dx}+z\frac{d{y}_1}{dx}=Q-P{y}_2$
$\Rightarrow y_1\frac{dz}{dx}+z\frac{d{y}_1}{dx}+P{y}_{1}z=Q$
$\Rightarrow y_1\frac{dz}{dx}+zQ=Q\Rightarrow y_1\frac{dz}{dx}=Q(1-z)$
$\Rightarrow \int \frac{dz}{1-z}=\int \frac{Q}{y_1}dx$
$\Rightarrow ln|z-1|=-\int \frac{Q}{y_1}dx+\lambda$
$\Rightarrow z=1+a{e}^{-\int \frac{Q}{y_1}dx}$