Question

Suppose $y=y(x)$ be the solution curve to the differential equation $\frac{d y}{d x}-y=2-e^{-x}$ such that $\displaystyle\lim _{x \rightarrow \infty} y(x)$ 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 $a-4 b$ is equal to ______

Answer 1

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