Practicing Success
Let $y_1$ and $y_2$ be the two solutions of the differential equation $\frac{d y}{d x}+P y=Q$, where $P$ and $Q$ are functions of $x$, Statement-1: The linear combination $a y_1+b y_2$ will be a solution of the differential equation, if $a+b=1$. Statement-2: The general solution of the differential equation is $y=y_1+C\left(y_1-y_2\right)$ |
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. Statement-1 is True, Statement-2 is False. Statement-1 is False, Statement-2 is True. |
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. |
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. |