I'm trying to find the solution of this differential equation;
$$y \cdot \frac{\partial{f}}{\partial{y}} -2x \cdot \frac{\partial{f}}{\partial{x}} = 2xy^4, \quad f(1,y) = \frac{1}{2}y^4 $$
But I think I went wrong somewhere because when I try to differentiate my solution I don't get the right answer! This is my solution,
I substituted $u = xy^2$ and $v=y$ then:
\begin{align*}
\frac{\partial{f}}{\partial{x}} &= y^2 \cdot \frac{\partial{f}}{\partial{u}}\\[10pt]
\frac{\partial{f}}{\partial{y}} &= 2xy \cdot \frac{\partial{f}}{\partial{u}} + \frac{\partial{f}}{\partial{v}}
\end{align*}
Then I get:
$$ \frac{\partial{f}}{\partial{v}} = 2xy^3 \implies f(v) = 2xy^3\cdot v + C(u)$$
\begin{align*}
f(x,y) &= 2xy^4 + C(xy^2)\\[5pt]
f(1,y) &= 2y^4 + C(y^2) = \frac{1}{2} y^2
\end{align*}
$$\implies C(y^2) = \frac{-3}{2}y^4 \implies C(xy^2) = \frac{-3}{2}x^2y^4
$$
$$ f(x,y) = 2xy^4 – \frac{3}{2}x^2y^4$$
Which apparently is wrong, what did I miss here?
Best Answer
You have two mistakes here. The first is when you write $$\frac{\partial{f}}{\partial{v}} = 2xy^3 \implies f(v) = 2xy^3\cdot v + C(u).$$ Because $x$ and $y$ are not independent from $u$ and $v$, you really need to do a full change of variables before integrating. You should get $$\frac{\partial f}{\partial v} = 2xy^3 = 2uv \implies f(v) = uv^2 + C(u).$$ Second, you also have an algebra error. When you wrote $$f(1,y) = 2y^4 + C(y^2) = \frac{1}{2} y^2 \implies C(y^2) = \frac{-3}{2}y^4$$ you combined a 4th degree term with a 2nd degree term. Make the first correction, and be careful with your algebra as you adjust the rest of your solution accordingly, and you should get to the right answer.