Question

Solve: $xdx+ydy+4y^3(x^2+y^2)dy=0$

• A. $1/2\ln (x^2+y^2)+y^4=C$
• B. $\ln (x^2+y^2)+y^4=C$
• C. $\ln (x^2+y^2)+x^4=C$
• D. $\ln (x^2+y^2)+y^2=C$

Answer 1

Answer: (A)

$1/2\ln (x^2+y^2)+y^4=C$

$xdx+ydy+4y^3(x^2+y^2)dy=0$
$\Rightarrow x+(4(x^2+y^2)y^3+y)\frac{dy}{dx}=0$ ...(1)
Let $R(x,y)=x$ and $S(x,y)=4(x^2+y^2)y^3+y$
This is not an exact equation, because $\frac{dR(x,y)}{dy}=0\ne 8x{y}^3=\frac{dS(x,y)}{dx}$
Find an integrating factor $\mu$ such that $\mu R(x,y)+\mu \frac{dy}{dx}S(x,y)=0$ is exact
This means $\frac{d}{dy}\left(\mu R(x,y)\right)=\frac{d}{dx}\left(\mu S(x,y)\right)$
$\Rightarrow x\frac{d\mu }{dy}=8x{y}^3\mu \Rightarrow \frac{\frac{d\mu }{dy}}{\mu }=8y^3\Rightarrow \log \mu =2y^4\Rightarrow \mu =e^{2y^4}$
Multiply both sides of (1) by $\mu$
$x{e}^{2y^4}+e^{2y^4}(4(x^2+y^2)y^3+y)\frac{dy}{dx}=0$
Let $P(x,y)=x{e}^{2y^4}$ and $Q(x,y)=e^{2y^4}(4(x^2+y^2)y^3+y)$
This is an exact equation, because $\frac{dP(x,y)}{dy}=8x{e}^{2y^4}{y}^3=\frac{dQ(x,y)}{dx}$
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
$f(x,y)=\int x{e}^{2y^4}dx=\frac{1}{2}{e}^{2y^4}{x}^2+g\left(y\right)$
Differentiating $f(x,y)$ w.r.t y
$\frac{df(x,y)}{dy}=\frac{d}{dy}(\frac{1}{2}{e}^{2y^4}{x}^2+g\left(y\right))=4e^{2y^4}{y}^3{x}^2+\frac{dg\left(y\right)}{dy}$
Substitute into $\frac{df(x,y)}{dy}=Q(x,y)$
$4e^{2y^4}{y}^3{x}^2+\frac{dg\left(y\right)}{dy}=e^{2y^4}(4(x^2+y^2)y^3+y)\Rightarrow \frac{dg\left(y\right)}{dy}=e^{2y^4}(4y^5+5)\Rightarrow g\left(y\right)=\int e^{2y^4}(4y^5+5)dy=\frac{1}{2}{e}^{2y^4}{y}^2$
Hence $1/2\ln (x^2+y^2)+y^4=C$