Question

Let $y=y_1(x)$ and $y=y_2(x)$ be two distinct solutions of the differential equation $\frac{dy}{dx}=x+y$, 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)

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