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Q.
The solution of differential equation $\frac{d y}{d x}-\frac{1}{x}+e^y \cdot x=0, y(1)=0$
Differential Equations
Solution:
$ \frac{ dy }{ dx }-\frac{1}{ x }=- e ^{ y } x \Rightarrow e ^{- y } \frac{ dy }{ dx }-\frac{1}{ x } e ^{- y }=- x$
$\text { put } e ^{- y }= z \Rightarrow \frac{ dz }{ dx }+\frac{1}{ x } \cdot z = x $
$\text { I.F. }= x$
solution $z \cdot x=\int x^2 d x+C \Rightarrow z x=\frac{x^3}{3}+c$
$e ^{- y } \cdot x =\frac{ x ^3}{3}+ c $
$e ^0 \cdot 1=\frac{1}{3}+ C \Rightarrow C =\frac{2}{3} $
$e ^{- y } \cdot x =\frac{ x ^3}{3}+\frac{2}{3} \Rightarrow 3 \cdot e ^{- y } x = x ^3+2$
$3 x =\left( x ^3+2\right) e ^{ y }$