2
$\begingroup$

I want to solve the differential equation $y'y^2=y+xy'$. I notice that the differential is not in the general form of first order linear equations which is $y'+P(x)y=Q(x)$. In fact the above equation is not linear.

I read that I have to switch the dependent and independent variable of the problem and then the problem will turn into linear first order differential equation. I don't quite understand the concept of switching dependent and independent variable, can someone explain?

$\endgroup$
  • 1
    $\begingroup$ Direct integration yields $y^3-3xy=3k$ for some constant $k$, hence $x=\frac{y^2}{3}-\frac{k}{y}$. $\endgroup$ – Omran Kouba Sep 16 '15 at 20:29
3
$\begingroup$

I'm not sure, what do you mean by switching variables, but if you simply need to solve it, here's what I'd do $$ y'y^2 = y + xy' \implies \left(\frac {y^3}3 \right)' = (xy)' \implies \frac {y^3}3 = xy + C $$ You can leave it as it is in the form of implicit function, or solve for $x$ $$ x = \frac {y^3 - 3C}{3y} $$

$\endgroup$
2
$\begingroup$

To follow on from the comment above. Two things to know $$ \dfrac{d}{dx} g(y) = \dfrac{dg}{dy}y' $$ And $$ f'(x)y + f(x)y' = (f(x)y)' $$ Always look for tricks like the above when solving (or trying to solve) odes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.