Question

The solution of the differential equation $(2x+10y^3)\frac{dy}{dx}+y=0$ is:

• A. $x+y=c{e}^{2x}$
• B. $y^2=2x^3+c$
• C. $x{y}^2=2y^5+c$
• D. $x(y^2+xy)=0$

Answer 1

The correct option is C $x{y}^2=2y^5+c$

$y+(10y^3+2x)\frac{dy}{dx}=0$ ...(1)
Let $R(x,y)=y$ and $S(x,y)=2x+10y^3$
This is not an exact equation, because $\frac{dR(x,y)}{dy}=1\ne 2=\frac{dS(x,y)}{dx}$
Find an integrating factor $\mu$ such that
$\mu R(x,y)+\mu \frac{dy}{dx}S(x,y)=0$ is exact
This means $\frac{d}{dy}\left(\mu R(x,y)\right)=\frac{d}{dx}\left(\mu S(x,y)\right)$
$\Rightarrow y\frac{d\mu }{dy}+\mu =2\mu \Rightarrow \frac{\frac{d\mu }{dy}}{\mu }=\frac{1}{y}\Rightarrow \log \mu =\log y\Rightarrow \mu =y$
Multiply both sides of (1) by $y$
$y^2+\left(2(x+5y^3)y\right)\frac{dy}{dx}=0$
Let $P(x,y)=y^2$ and $Q(x,y)=2y(x+5y^2)$
This is exact equation, because $\frac{dP(x,y)}{dy}=2y=\frac{dQ(x,y)}{dx}$
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
$f(x,y)=\int y^{2}dx=y^{2}x+g\left(y\right)$
where $g\left(y\right)$ is an arbitrary function of y
Differentiate $f(x,y)$ w.r.t y
$\frac{df(x,y)}{dy}=\frac{d}{dy}(y^{2}x+g\left(y\right))=2yx+\frac{dg\left(y\right)}{dy}$
substitute into $\frac{df(x,y)}{dy}=Q(x,y)$
$2yx+\frac{dg\left(y\right)}{dy}=2y(x+5y^3)\Rightarrow \frac{dg\left(y\right)}{dy}=10y^4\Rightarrow g\left(y\right)=\int 10y^{4}dy=2y^5$
hence $2y^5+x{y}^2=c$