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Solve $2y+(x^2 y + 1)xy'=0$. The book that contains this problem suggests a way of solving such differential equations - do a substitution $y=z^m$, $dy=m z^{m-1}dz$ with such $m$ that the differential equation becomes homogenous. Here $m=-2$ works. However if I substitute $y=z^{-2}$ I lose all information about solutions with $y \leq 0$.

Let's say I do this substitution nonetheless.

$$2z^{-2} + (x^2z^{-2}+1)x(-2)z^{-3}z'=0$$

$$z^{-2}dx = (x^3z^{-5} +x z^{-3})dz$$

Now it's homogenous. Substitute $t=\frac{x}{z}, dx=tdz+zdt$ $$tdz+zdt = (t^3+t)dz$$ $$\frac{dz}{z}=\frac{dt}{t^3}$$ Note that we lost $t=x=0$, but it's not a solution. $$\ln{|z|} = -\frac{1}{2t^2} + C,$$ Substitute back: $$\ln{\frac{1}{\sqrt{y}}} = -\frac{1}{x^2 y} + C.$$

Check whether $y=0$ is a solution - it is. But the book containing this problem says the solutions are $x^2 y \ln{Cy}=1$ and $y=0$. It allows for $y$ to be negative, but our solution does not. How to go about it?

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1 Answer 1

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Making $z = \frac{1}{x^2y}$ and substituting we obtain

$$ 2y+(x^2y+1)x y'= 0 \equiv 2z+x(1+z)z'=0 $$

and now

$$ \frac{(z+1)}{2z}dz = - \frac{dx}{x} $$

etc.

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  • $\begingroup$ How did you know to make this substitution? $\endgroup$
    – CrabMan
    Apr 20, 2018 at 19:38
  • $\begingroup$ Mathematics involves a lot of imagination. $\endgroup$
    – Cesareo
    Apr 20, 2018 at 20:26

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