# Solution to a exact differential equation

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

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

• where is an "equation"? – haqnatural Oct 19 '16 at 20:45
• I forgot so sorry i just added – user96369 Oct 19 '16 at 20:48
• 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 $C_2$. – Jacky Chong Oct 19 '16 at 20:48
• Basically, you are almost done. – Jacky Chong Oct 19 '16 at 20:48
• in $f$ , you forgot a minus. – hamam_Abdallah Oct 19 '16 at 20:57

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