Let y_1 and y_2 be the two solutions of the differential equation (d…

It is given that $y_1$ and $y_2$ are solutions of the differential equation

$\frac{d y}{d x}+P y=Q$        .....(i)

∴  $\frac{d y_1}{d x}+P y_1=Q$        .....(ii)

and, $\frac{d y_2}{d x}+P y_2=Q$        .....(iii)

On subtracting (ii) from (i) and (iii) from (ii), we obtain

$\frac{d}{d x}\left(y-y_1\right)+P\left(y-y_1\right)=0$ and, $\frac{d}{d x}\left(y_1-y_2\right)+P\left(y_1-y_2\right)=0$

$\Rightarrow \frac{d}{d x}\left(y-y_1\right)=-P\left(y-y_1\right)$ and, $\frac{d}{d x}\left(y_1-y_2\right)=-P\left(y_1-y_2\right)$

$\Rightarrow \frac{\frac{d}{d x}\left(y-y_1\right.}{\frac{d}{d x}\left(y_1-y_2\right)}=\frac{y-y_1}{y_1-y_2}$

$\Rightarrow \frac{d\left(y-y_1\right)}{y-y_1}=\frac{d\left(y_1-y_2\right)}{y_1-y_2}$

On integrating, we get

$\log \left(y-y_1\right)=\log \left(y_1-y_2\right)+\log C$

$\Rightarrow y-y_1=C\left(y_1-y_2\right)$

$\Rightarrow y=y_1+C\left(y_1-y_2\right)$

Hence, $y=y_1+C\left(y_1-y_2\right)$ is the general solution of the given differential equation.

So, statement-2 is true.

Now,

$y =y_1+C\left(y_1-y_2\right)$

$\Rightarrow y =(1+C) y_1+(-C) y_2$

$\Rightarrow y =a y_1+b y_2$, where $a=1+C$ and $b=-C$ i.e. $a+b=1$

Hence, $y=a y_1+b y_2$ is the general solution iff $a+b=1$

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