Question

Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{dy}{dx}-y=2-e^{-x}$ such that $\underset{x\to \infty }{\lim }y(x)$ is finite. If $a$ and $b$ are respectively the $x$-and $y$ - intercepts of the tangent to the curve at $x=0$, then the value of $a-4b$ is equal to ______

Answer 1

Answer: 3

$\frac{dy}{dx}-y=2-e^{-x}$
I.F. $=e^{-\int dx}=e^{-x}$
$herefore$ solution of D.E
$y\cdot e^{-x}=\int \left(2e^{-x}-e^{-2x}\right)dx$
$\Rightarrow y=-2+\frac{e^{-x}}{2}+C\cdot e^x$
$\because \underset{x\to \infty }{\lim }y$ is finite
$\therefore \underset{x\to \infty }{\lim }\left(-2+\frac{e^x}{2}+C\cdot e^x\right)\to$ finite
This is possible only when $C=0$
$\therefore y=y(x)=-2+\frac{e^{-x}}{2}$
$\frac{dy}{dx}=-\frac{1}{2}{e}^{-x}$
${\frac{dy}{dx}|}_{x=0}=-\frac{1}{2}=m,y(0)=-2+\frac{1}{2}=\frac{-3}{2}$
$\therefore$ equation of tangent
$y+\frac{3}{2}=-\frac{1}{2}(x-0)$
$\Rightarrow x+2y=-3$
$a=-3,b=\frac{-3}{2}$
$a-4b=-3+6=3$