0
$\begingroup$

Here i have an exact differential equation which is

$\frac{xdx}{(x^2+y^2)^{3/2}} +\frac{ydy}{(x^2+y^2)^{3/2}}=0$

My solution is that:

$f=\int{Mdx} +g(y)$ Here M is from $Mdx+ Ndy$

$f=\frac{1}{2\sqrt{x^2+y^2}}+ g(y)$

Since $\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}$

$\frac{y}{(x^2+y^2)^{3/2}}+g'(y)= \frac{y}{(x^2+y^2)^{3/2}}$

We get $g(y)=c$ However answer to this question is

$x^2+y^2=c^2$

$\endgroup$
  • $\begingroup$ where is an "equation"? $\endgroup$ – haqnatural Oct 19 '16 at 20:45
  • $\begingroup$ I forgot so sorry i just added $\endgroup$ – user96369 Oct 19 '16 at 20:48
  • $\begingroup$ Since you have shown $f(x, y) = \frac{1}{2}(x^2+y^2)^{-1/2}+C$, then the solution to the exact equation is implicitly given by $f(x, y) = C_1$ for come constant $C_1$. Thus, moving this around get you $\sqrt{x^2+y^2} = C_2$ for some other constant $C_2$. $\endgroup$ – Jacky Chong Oct 19 '16 at 20:48
  • $\begingroup$ Basically, you are almost done. $\endgroup$ – Jacky Chong Oct 19 '16 at 20:48
  • $\begingroup$ in $f$ , you forgot a minus. $\endgroup$ – hamam_Abdallah Oct 19 '16 at 20:57
1
$\begingroup$

$$(x^{ 2 }+y^{ 2 })^{ 3/2 }\cdot \left( \frac { xdx }{ (x^{ 2 }+y^{ 2 })^{ 3/2 } } +\frac { ydy }{ (x^{ 2 }+y^{ 2 })^{ 3/2 } } \right) =0\cdot (x^{ 2 }+y^{ 2 })^{ 3/2 }\\ xdx+ydy=0\\ \frac { 1 }{ 2 } d\left( { x }^{ 2 }+{ y }^{ 2 } \right) =0\\ { x }^{ 2 }+{ y }^{ 2 }=c$$

$\endgroup$

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.