Let y_1 and y_2 be the solutions of the differential equation (d y)/(…

It is given that $y_1$ and $y_2$ are solutions of

$\frac{d y}{d x}+P y=Q $

∴   $\frac{d y_1}{d x}+P y_1=Q$ and $\frac{d y_2}{d x}+P y_2=Q$

Now,

$\frac{d y_2}{d x}+P y_2=Q$ and $y_2=y_1 z$

$\Rightarrow \frac{d}{d x}\left(y_1 z\right)+P y_1 z=Q$

$\Rightarrow \left(\frac{d y_1}{d x}+P y_1\right) z+y_1 \frac{d z}{d x}=Q$

$\Rightarrow Q z+y_1 \frac{d z}{d x}=Q$

$\Rightarrow \frac{d z}{d x}+\frac{Q}{y_1} z=\frac{Q}{y_1}$            ......(i)

This is a linear differential equation with integrating factor $=e^{\int \frac{Q}{y_1} d x}$

Multiplying both sides of (i) by integrating factor and integrating with respect to $x$, we get

$z e^{\int \frac{Q}{y_1} d x}=\int\left\{\frac{Q}{y_1} e^{\int \frac{Q}{y_1} d x}\right\} d x+C$

$\Rightarrow z e^{\int \frac{Q}{y_1} d x}=e^{\int \frac{Q}{y_1} d x}+C \Rightarrow z=1+C e^{-\int \frac{Q}{y_1} d x}$

So, statement-2 is true.

∴  $y_2=y_1 z$

$\Rightarrow y_2=\left\{1+C e^{-\int \frac{Q}{y_1} d x}\right\} y_1 \Rightarrow \frac{y_2-y_1}{y_1}=C e^{-\int \frac{Q}{y_1} d x}$

So, statement- 1 is also true. Also, statement-2 is a correct explanation for statement- 1.