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Q.
Solution of the differential equation $\frac{d x}{d y}-\frac{x \ln x}{1+\ln x}=\frac{e^y}{1+\ln x}$ if $y(1)=0$ is
Differential Equations
Solution:
$(1+\ln x) \frac{d x}{d y}-x \ln x=e^y$
$t = ye ^{ y }+ e ^{ y } \cdot c $
$x \ln x = ye ^{ y }+ e ^{ y } \cdot c $
$0= c $
$\Rightarrow x \ln x = ye ^{ y }$
$\ln x ^{ x }= y e ^{ y } $
$x ^{ x }= e ^{ ye ^{ y }}$
