Question
Find the differential equation by eliminating arbitrary constants from the relation $y = (c_1 + c_2x)e^x$

Answer

$ y=\left(c_1+c_2 x\right) e^x$
$\therefore y e^{-x}=c_1+c_2 x $
Differentiating w.r.t. $x$, we get
$ y\left(- e ^{-x}\right)+ e ^{-x} \frac{ d y}{ d x}=0+ c _2$
$\therefore e ^x\left(\frac{ d y}{ d x}-y\right)= c _2 $
Again, differentiating w.r.t. $x$, we get
$ e ^{-x}\left(\frac{ d ^2 y}{ d x^2}-\frac{ d y}{ d x}\right)- e ^{-x}\left(\frac{ d y}{ d x}-y\right)=0$
$\therefore e ^{-x}\left(\frac{ d ^2 y}{ d x^2}-\frac{ d y}{ d x}-\frac{ d y}{ d x}+y\right)=0$
$\therefore \frac{ d ^2 y}{ d x^2}-2 \frac{ d y}{ d x}+y=0 $

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