Question

The solution of the differential equation $\frac{dy}{dx}=\frac{y^3}{e^{2x}+y}$ is given by:

• A. $y^2+2(c+\log y)e^{2x}=0$
• B. $x^2+2(c+\log y)e^{2x}=0$
• C. $e^{2x}{x}^2+y^2{e}^{2y}+c=0$
• D. None of these

Answer 1

The correct option is A $y^2+2(c+\log y)e^{2x}=0$

$\frac{dy}{dx}=\frac{y^3}{e^{2x}+y}-y^3+(e^{2}x+y)y^{\prime }\left(x\right)=0$ ...(1)
Let $R(x,y)=-y^3$ and $S(x,y)=e^{2}x+y$
This is not an exact equation. because
$\frac{dR(x,y)}{dy}=-{3y}^2\ne e^2=\frac{dS(x,y)}{dx}$
Find an integrating factor $\mu$ such that
$\mu R(x,y)+\mu \frac{dy}{dx}S(x,y)=0$ is exact
This means $\frac{d}{dx}\left(\mu R(x,y)\right)=\frac{d}{dx}\left(\mu \left(S(x,y)\right)\right)$
$\therefore -y^3\frac{d\mu }{dy}-{3y}^2\mu =e^2\mu \Rightarrow \frac{\frac{d\mu }{dy}}{\mu }=-\frac{{3y}^2+e^2}{y^3}$
$\Rightarrow \log \mu =\frac{e^2}{{2y}^2}-3\log y\Rightarrow \mu =\frac{e^{\frac{e^2}{{2y}^2}}}{y^3}$
Multiplying both sides of (1) by $\mu$
$-e^{\frac{e^2}{2y^2}}+\left(\frac{e^{\frac{e^2}{{2y}^2}}(e^{2}x+y)}{y^3}\right)\frac{dy}{dx}=0$
Let $P(x,y)=-e^{\frac{e^2}{{2y}^2}}$ and $Q(x,y)=\frac{e^{\frac{e^2}{{2y}^2}}(e^{2}x+y)}{y^3}$
This is an exact equation because
$\frac{dP(x,y)}{dy}=\frac{e^{\frac{e^2}{{2y}^2}+2}}{y^3}=\frac{dQ(x,y)}{dx}$
Define $f(x,y)$ such that
$\frac{df(x,y)}{dx}=p(x,y)$ and $\frac{df(x,y)}{dy}=Q(x,y)$
Then the solution will be given by $f(x,y)=c$
Integrate $\frac{df(x,y)}{dx}$ w.r.t x
$f(x,y)=\int -e^{\frac{e^2}{{2y}^2}}dx=-e^{\frac{e^2}{{2y}^2}}+g\left(y\right)$
Differentiate $f(x,y)$ w.r.t y to find $g\left(y\right)$
$\frac{df(x,y)}{dy}=\frac{d}{dy}(-e^{\frac{e^2}{{2y}^2}}x+g\left(y\right))=\frac{e^{\frac{e^2}{{2y}^2}}x}{y^3}+\frac{dg\left(y\right)}{dy}$
Substitute into $\frac{df(x,y)}{dy}=Q(x,y)$
$\frac{e^{\frac{e^2}{{2y}^2}}x}{y^3}+\frac{dg\left(y\right)}{dy}=\frac{e^{\frac{e^2}{{2y}^2}}(e^{2}x+y)}{y^3}$
Solve for $\frac{dg\left(y\right)}{dy}$
$\frac{dg\left(y\right)}{dy}=\frac{e^{\frac{e^2}{2y^2}}}{y^2}\Rightarrow \int \frac{dg\left(y\right)}{dy}=\int \frac{e^{\frac{e^2}{{2y}^2}}}{y^2}\Rightarrow y^2+2(c+\log y)e^{2x}=0$