1
$\begingroup$

Justify that a differential equation of the form: $$[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$$ , where $f(x^2+y^2)$ is an arbitrary function of $(x^2+y^2)$, is not an exact differential equation and $1/(x^2+y^2)$ is an integrating factor for it. Hence, solve this differential equation for $$f(x^2+y^2)=(x^2+y^2)^2$$

I am unable to solve after making it an exact equation, having difficulty in the integration of this question.

$\endgroup$
1
  • 1
    $\begingroup$ Please refrain from posting questions as pictures: use MathJax to format questions, and also add your working. $\endgroup$ Oct 29, 2020 at 13:31

1 Answer 1

1
$\begingroup$

$$(y+xf(x^2+y^2))dx+(yf(x^2+y^2)-x)dy=0$$ $$f(x^2+y^2))(xdx+ydy)+ydx-xdy=0$$ $$\dfrac 12f(x^2+y^2))(d(x^2+y^2))+ydx-xdy=0$$ Divide by $x^2+y^2$: $$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}+\dfrac {ydx-xdy}{x^2+y^2}=0$$ Note that: $$d \arctan (y/x)=\dfrac {xdy-ydx}{x^2+y^2}$$ $$\dfrac {f(x^2+y^2)d(x^2+y^2)}{2(x^2+y^2)}-d(\arctan (y/x)) =0$$ Integrate for the given function $f$.

$\endgroup$

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.