Solve Exact Equation: x dy-(2xe^x-y+6x^2) dx = 0

[misogynisticfeminist]

I've got a problem with exact equations here, the question I've got is,

$$x \frac {dy}{dx} = 2xe^x-y+6x^2$$

sp, i put it in the form,

$$x dy-(2xe^x-y+6x^2) dx = 0$$

$$x dy+(-2xe^x+y-6x^2) dx = 0$$

the equation would be exact as,

$$\frac {\partial M}{\partial x}=1$$

$$\frac {\partial N}{\partial y} =1$$

But when I integrate M wrt. y and N wrt x I get totally different answers. So which one do I follow? Thanks.

: )

 

[einstone]

Since the equation is exact, there is a function F(x,y) such that
M = The derivative of F with respect to y ... (1)

N= The derivative of F with respect to x. ...(2)
So, F = The integral of M with respect to y ( keeping x constant) + a function of x (which I'll call g(x)).
Differentiating with respect to x,
N = d/dx ( Int.Mdy) + g'(x). Now g can be ( in principle) found out.

Are you sure you included g in the integration?
I'm,with great respect,
Einstone.

 

[misogynisticfeminist]

thanks for the help, i forgot to include the g(x) as you have said. I don't think I'll start another thread but I've now currently got a problem. It has something to do with the linear equation.

$$x^-^4 \frac {dy}{dx} 4x^-^5 y = xe^x$$

I don't know how to simplify this to,

$$\frac {d}{dx} (x^-^4 y) = xe^x$$

Thanks...

(I'm now at my friend's house and the book is not with me, so I'll dig out the original question soon.)

 

[Zurtex]

Erm, I'm not convinced you wrote this out correctly (I'm guessing there should be - between the dy/dx and the 4x^-5 but if you did:

$$x^{-4} \frac {dy}{dx} 4x^{-5} y = xe^x$$

$$\frac{dy}{dx} \frac{1}{x^9} y = xe^x$$

$$y\frac{dy}{dx} = x^{10} e^x$$

$$\int y \frac{dy}{dx} dx = \int x^{10} e^x$$

Then all you need to do is use by-parts a lot. But assuming you meant there to be a minus there:

$$x^{-4} \frac {dy}{dx} - 4x^{-5} y = xe^x$$

Notice the LHS takes the form d/dx (uv) = uv'+u'v, where u=x^-4 and v=y. So rewriting:

$$\frac{d}{dx} \left( x^{-4} y \right) = xe^x$$

 

[misogynisticfeminist]

I'm sorry, there was a mistake, it should be,

$$x^-^4 \frac {dy}{dx} - 4x^-^5y =xe^x$$

there should be a minus sign...

I've finally understood, thanks a lot...

: )