# Q18.    The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is            (A)    $xe^y + x^2 = C$            (B)    $xe^y + y^2 = C$            (C)    $ye^x + x^2 = C$            (D)    $ye^y + x^2 = C$

G Gautam harsolia

Given equation is
$e^x dy + (y e^x + 2x) dx = 0$
we can rewrite it as
$\frac{dy}{dx}+y=-2xe^{-x}$
It is $\frac{dy}{dx}+py=Q$ type of equation where $p = 1 \ and \ Q = -2xe^{-x}$
Now,
$I.F. = e^{\int p dx }= e^{\int 1dx}= e^x$
Now, the general solution is
$y(I.F.) = \int (Q\times I.F.)dx+C$
$y(e^x) = \int (-2xe^{-x}\times e^x)dx+C\\ ye^x= \int -2xdx + C\\ ye^x=- x^2 + C\\ ye^x+x^2 = C$
Therefore, (C) is the correct answer

Exams
Articles
Questions