Question

Find all solutions of y'-(1+x)/3x*y=y^4.

Answer 1

@SithsAndGiggles

Answer 2

v=y^-3=1/y^3
y^3=1/v
y=(1/v)^(1/3)
(1/3)(1/v)^(-2/3)*dv/dx-(1+x)/3x*(1/v)^(1/3)=(1/v)^(4/3)
(v^2/3)dv/dx-(1+x)/3x*(1/v)=(1/v)^4
dv/dx-(1+x)/3x*3/v^3=(1/v^4)(3/v^2)
dv/dx-(1+x)/(xv^3)=3/v^6
u(x)=e^(-ln abs(x)-x)=1/x*e^-x=1/xe^x

Answer 3

Is the second term \(\dfrac{1+x}{3x}y\) or \(\dfrac{1+x}{3}xy\) ?

Answer 4

the first one you wrote.

Answer 5

\[\begin{align*}
y'-\frac{1+x}{3x}y&=y^4\\
y^{-4}y'-\frac{1+x}{3x}y^{-3}&=1\\
-\frac{1}{3}v'-\frac{1+x}{3x}v&=1&\text{letting }v=y^{-3}\text{, so }v'=-3y^{-4}y'\\
v'+\frac{1+x}{x}v&=-3\end{align*}\]
Integrating factor:
\[\mu(x)=\exp\left(\int \frac{1+x}{x}~dx\right)=\exp\left(\ln|x|+x\right)=xe^x\]
\[\begin{align*}
xe^xv'+(xe^x)\frac{1+x}{x}v&=-3xe^x\\
xe^xv'+(e^x+xe^x)v&=-3xe^x\\
\frac{d}{dx}\left[xe^xv\right]&=-3xe^x\\
xe^xv&=-3\int xe^x~dx\\
xe^xy^{-3}&=-3\left(xe^x-\int e^x~dx\right)&\text{(integrating by parts)}\\
xe^xy^{-3}&=-3xe^x+3e^x+C\\
y^{-3}&=-3+\frac{3}{x}+\frac{C}{xe^x}
\end{align*}\]

Answer 6

Wait a minute.

Answer 7

I have to get going, sorry... I should be back on later. I provided an implicit solution, but solving for an explicit one shouldn't be too hard.

Answer 8

That's fine. I can solve for the explicit solution. Thanks.