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So I was presented with this equation in my textbook, currently studying multi variable calculus:

$$yf^{'}_x(x,y) - xf^{'}_y(x,y) = f(x,y)$$

Using the substitution:

$$x^2 + y^2 = u$$ $$e^{{-x^2}{/2}} = v$$

I get the equation (which is correct):

$$v f^{'}_v + f = 0$$

My next task is to find the solution, $f(x,y)$ that satisfies $f(0, y) = y^2$. The correct answer should be $f(x,y) = (x^2 + y^2)e^{{-x^2}{/2}}$.

But here I am stuck, I have solved similar tasks but I don't know how to handle the $f$. For example $v f^{'}_v = x$ or similar would be easy to solve.

Lukas Arvidsson
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1 Answers1

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Hint: the left-hand side equals $(v f)'_v$ (derivative of a product).

Hans Lundmark
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  • Thank you, not sure exactly how you mean, if you have time I would love a bit more explanation :) – Lukas Arvidsson Oct 04 '18 at 14:03
  • I assume you know the product rule $(fg)'=f'g+fg'$ from single-variable calculus, right? Just apply that to the product $v \cdot f(u,v)$ with the partial derivative $\partial/\partial v$ instead of the ordinary derivative. It works the same, since the partial derivative is just an ordinary derivative once you have fixed a value of $u$. – Hans Lundmark Oct 04 '18 at 14:06
  • And you have also probably seen how to solve first-order ordinary differential equations using an integrating factor? This is the same type of problem. – Hans Lundmark Oct 04 '18 at 14:08