Bernoulli Equation

[cartman]

I have been trying to solve this equation all week from my math textbook. Unfortunately, there is no solutions manual so I do not know what steps to take.

\(\displaystyle xy^2y'=x^3+y^2\)

This problem is in the substitution section of the textbook so I am assuming that it needs to be solved using the Bernoulli Equation method due to \(\displaystyle y\) being to a higher power.

Of course, I can reduce it to \(\displaystyle y'-\frac{1}{x}=x^2y^{-2}\) and then substitute in \(\displaystyle v=y^{3}\), which gives me

\(\displaystyle y'=3x^{2}+\frac{3v^{\frac{2}{3}}}{x}\), but \(\displaystyle v\) is still not to the first power. Any suggestions on how to solve this?

I know that the final result is supposed to work out to be \(\displaystyle y^3=3x^3(C+ln(\abs{x}))\).

Thank you

 

[galactus]

I think you have a typo. Should be:

\(\displaystyle xy^{2}\frac{dy}{dx}=x^{3}+\boxed{y^{3}}\)

If that is a book typo, then no wonder you struggled with it.

It's not a Bernoulli because it is not of the form \(\displaystyle \frac{dy}{dx}+P(x)y=f(x)y^{n}\)

Instead, make the sub \(\displaystyle y=ux, \;\ dy=udx+xdu\)

Then, it becomes:

\(\displaystyle u^{2}x^{3}(udx+xdu)-(x^{3}+u^{3}x^{3})dx=0\)

\(\displaystyle u^{2}xdu-dx=0\)

\(\displaystyle u^{2}du-\frac{dx}{x}=0\)

\(\displaystyle \int u^{2}du-\int\frac{1}{x}dx=0\)

\(\displaystyle \frac{u^{3}}{3}-ln|x|=C\)

Resub \(\displaystyle u=\frac{y}{x}\) and let \(\displaystyle C_{1}=3C\)

\(\displaystyle -3x^{3}ln|x|+y^{3}=C_{1}x^{3}\)

\(\displaystyle \boxed{y^{3}=C_{1}x^{3}+3x^{3}ln(x)}\)

 

[cartman]

Wow, you are right, I copied it from the book incorrectly. Great catch. I never would have solved it. Thank you.