Question

Let $y = y _1( x )$ and $y = y _2( x )$ be two distinct solutions of the differential equation $\frac{d y}{d x}=x+y$, with $y _1(0)=0$ and $y _2(0)=1$ respectively. Then, the number of points of intersection of $y=y_1(x)$ and $y=y_2(x)$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

$ \frac{d y}{d x}=x+y \Rightarrow \frac{d y}{d x}-y=x$
$ \therefore \text { solution is } y e^{-x}=\int x e^{-x} d x $
$ \Rightarrow y e^{-x}=-x e^{-x}-e^{-x}+c $
$\Rightarrow y=-x-1+c e^x $
$ y_1(0)=0 \Rightarrow c=1 $
$ \therefore y_1=-x-1+e^x ....$(1)
$ y _2(0)=1 \Rightarrow c =2$
$ \therefore y_2=-x-1+2 e^x .....$(2)
Now $y_2-y_1=e^x>0 \therefore y_2 \neq y_1$
$\therefore$ Number of points of intersection of $y _1 \& y _2$ is zero.