Question

Solution of the differential equation
$\left(e^{x^2}+e^{y^2}\right) y \frac{d y}{d x}+e^{x^2}\left(x y^2-x\right)=0, \text { is } $

• A. $e ^{ x ^2}\left( y ^2-1\right)+ e ^{y^2}= C$
• B. $e ^{y^2}\left( x ^2-1\right)+ e ^{ x ^2}= C$
• C. $e ^{ y ^2}\left( y ^2-1\right)+ e ^{ x ^2}= C$
• D. $e^{x^2}(y-1)+e^{y^2}=C$

Answer 1

Answer: (A)

$y ^2= t ; 2 y \frac{ dy }{ dx }=\frac{ dt }{ dx }$; Hence the differential equation becomes
$\left(e^{x^2}+e^t\right) \frac{d t}{d x}+2 e^{x^2}(x t-x)=0 $
$e^{x^2}+e^t+2 e^{x^2} \cdot x(t-1) \frac{d x}{d t}=0$
$\text { put } e^{x^2}=z ; e^{x^2} \cdot 2 x \frac{d x}{d t}=\frac{d z}{d t} $
$z+e^t+\frac{d z}{d t}(t-1)=0$
$\frac{ dz }{ dt }+\frac{ z }{( t -1)}=-\frac{ e ^{ t }}{( t -1)} ; \quad \text { I.F. }= e ^{\int \frac{ dt }{ t -1}}= e ^{\ln ( t -1)}= t -1 $
$z ( t -1)=-\int\left( e ^{ t }\right) dt $
$z ( t -1)=- e ^{ t }+ C$
$e ^{ x ^2}\left( y ^2-1\right)=- e ^{ y ^2}+ C$
$ e ^{ x ^2}\left( y ^2-1\right)+ e ^{ y ^2}=\text { C }$