Exact Equation

• September 11th 2010, 09:49 PM
staevobr
Exact Equation
$(e^x \sin{y} -3x^2) \,dx +(e^x \cos{y} + \frac{y^{-2/3}}{3}) \, dy = 0$

$(e^x \sin{y} -3x^2) +(e^x \cos{y} + \frac{y^{-2/3}}{3}) \frac{dy}{dx} = 0
$

$M = e^x \sin{y} -3x^2 \,, N = e^x \cos{y} + \frac{y^{-2/3}}{3}$

$\frac{\partial M}{\partial y} = e^x \cos{y}$

$\frac{\partial N}{\partial x} = e^x \cos{y}$

Since
$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$,
there exists an F such that
$\frac{d}{dx}F(x, y) = 0$

$F = \int{M dx}$
$F = \int{e^x \sin{y} -3x^2 dx} = e^x \sin{y} -x^3 + h(y)$

$\frac{\partial F}{\partial y} = N$

$e^x \cos{y} + \frac{y^{-2/3}}{3} = e^x \sin{y} -x^3 + h'(y)$

I can't find any way to get past this point. Is there any possible function $h'(y)$ that would satisfy this equality? Or have I made an error?
• September 11th 2010, 09:58 PM
Prove It
Your working is a little off...

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

that means

$\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$.

So

$\frac{\partial F}{\partial x} = e^x\sin{y} - 3x^2$

$F = \int{e^{x}\sin{y} - 3x^2\,dx}$

$= e^x\sin{y} - x^3 + g(y)$.

$\frac{\partial F}{\partial y} = e^x\cos{y} + \frac{y^{-\frac{2}{3}}}{3}$

$F = \int{e^x\cos{y} + \frac{y^{-\frac{2}{3}}}{3}\,dy}$

$= e^x\sin{y} + y^{\frac{1}{3}} + h(x)$.

So we have at the same time...

$F = e^x\sin{y} - x^3 + g(y)$ and $F = e^x\sin{y} + y^{\frac{1}{3}} + h(x)$.

This means $g(y) = y^{\frac{1}{3}} + C_1$ and $h(x) = -x^3 + C_2$.

Putting everything together gives

$F = e^x\sin{y} - x^3 + y^{\frac{1}{3}} + C$ where $C = C_1 + C_2$.
• September 11th 2010, 10:08 PM
staevobr
Thanks.