Question

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

• 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)

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