Question

The solution of the differential equation $(e^{x^2}+e^{y^2})y\frac{dy}{dx}+e^{x^2}(x{y}^2-x)=0$ is

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

Answer 1

The correct option is A $e^{x^2}(y^2-1)+e^{y^2}=C$

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