Question

Solve y'=(xy^2+2y^3)/(x^3+x^2*y+xy^2) implicitly.

Answer 1

aren't you getting bored of doing so many homogeneous equations ?

Answer 2

It's good practice :)

Answer 3

Here's the work:
y=ux
u=y/x
y'=u'x+u
u'x+u=(u^2+2u^3)/(u^2+u+1)
Okay, I will not waste time on posting all my work, it's just too much. I got the answer (y-x)^3(y+x)=cy^2*x^3 but the answer in the book is (x-y)^3*(x+y)=cy^2(x^4). I just don't know which answer is correct. Wolphfram sometimes don't have the clear answer as well.

Answer 4

\[y'=\frac{xy^2+2y^3}{x^3+x^2y+xy^2}~~\iff~~xv'+v=\frac{v^2x^3+2v^3x^3}{x^3+vx^3+v^2x^3}\]
Cancelling \(x^3\) gives
\[xv'+v=\frac{v^2+2v^3}{1+v+v^2}\]
Simplifying,
\[\begin{align*}xv'+v&=\frac{v^2+2v^3}{1+v+v^2}\\\\
xv'&=\frac{v^2+2v^3}{1+v+v^2}-v\frac{1+v+v^2}{1+v+v^2}\\\\
xv'&=\frac{v^2+2v^3-v-v^2-v^3}{1+v+v^2}\\\\
xv'&=\frac{v^3-v}{1+v+v^2}\\\\
\frac{v^2+v+1}{v(v+1)(v-1)}~dv&=\frac{dx}{x}
\end{align*}\]

Answer 5

Yes, that's what I got.

Answer 6

Okay, so partial fraction to deal with the left side:
\[\begin{align*}
\frac{v^2+v+1}{v(v+1)(v-1)}&=\frac{A}{v}+\frac{B}{v+1}+\frac{C}{v-1}\\\\
v^2+v+1&=A(v^2-1)+B(v^2-v)+C(v^2+v)\\
&=(A+B+C)v^2+(C-B)v-A
\end{align*}\]
Giving \(A=-1\), \(B=\dfrac{1}{2}\), \(C=\dfrac{3}{2}\).

So after integrating you would have
\[-\ln|v|+\frac{1}{2}\ln|v+1|-\frac{3}{2}\ln|v-1|=\ln|x|+C\]

Answer 7

Why do you have -3/2? I have 3/2?

Answer 8

You're right, that's a typo. Should be positive.

Answer 9

\[-\frac{1}{2}\left(2\ln|v|-\ln|v+1|-3\ln|v-1|\right)=-\frac{1}{2}\ln\left|\frac{v^2}{v^2-1}\right|=\ln\left|\sqrt{\frac{v^2-1}{v^2}}\right|=\ln|x|+C\]

Answer 10

...+C

Answer 11

Oh, so you factored out -1/2, I multiplied by 2 for the whole equation.

Answer 12

Right, same difference!

So we can reduce this to
\[\sqrt{\frac{v^2-1}{v^2}}=Cx\]
Back-substitute:
\[\sqrt{1-\frac{1}{v^2}}=\sqrt{1-\frac{x^2}{y^2}}=Cx\]