Integrating factor in Exact Equations

[alane1994]

Ok, so I have this differential equation.

\[(3x^2y+2xy+y^3)+(x^2+y^2)y\prime=0\]

First I needed to check to see if it is exact.

\(M=3x^2y+2xy+y^3\)
\(N=x^2+y^2\)

\(\dfrac{\partial M}{dy}(3x^2y+2xy+y^3)=3x^2+2x+3y^2\)

\(\dfrac{\partial N}{dx}(x^2+y^2)=2x+0\)

For the integrating factor, I believe it is in the form,

\[\dfrac{M_y(x,y)-N_x(x,y)}{N(x,y)}\mu\]

And so, I would have,

\(\dfrac{3x^2+2x+3y^2-2x}{x^2+y^2}\mu = \dfrac{3x^2+3y^2}{x^2+y^2}\mu=\dfrac{d\mu}{dx}\)

Now... I feel as though I am wrong in this. Could someone look and tell me if I am on the right track, and if I am off guide me back to the path of righteousness?

 

[Opalg]

And so, I would have,

\(\dfrac{3x^2+2x+3y^2-2x}{x^2+y^2}\mu = \dfrac{3x^2+3y^2}{x^2+y^2}\mu=\dfrac{d\mu}{dx}\)

Now... I feel as though I am wrong in this. Could someone look and tell me if I am on the right track, and if I am off guide me back to the path of righteousness?

You seem to be on the right track, but you need to go a little bit further, by noticing that $\dfrac{3x^2+3y^2}{x^2+y^2} = \dfrac{3(x^2+y^2)}{x^2+y^2} = 3.$ (Happy)

 

[alane1994]

\(\dfrac{d\mu}{dx}=3\mu\)

Would \(\mu(x)=0\)?

 

[Opalg]

\(\dfrac{d\mu}{dx}=3\mu\)

Would \(\mu(x)=0\)?

No. Think again (think exponential).

 

[Chris L T521]

\(\dfrac{d\mu}{dx}=3\mu\)Would \(\mu(x)=0\)?

Sure, but things would be boring if that was the only solution. Instead, try solving the ODE by separation of variables.

EDIT: Ninja'd by Opalg.

 

[alane1994]

I feel foolish, I am unsure how to get what is required.
I typed it into wolfram and got,
\(\mu(x)=e^{3x}\)

 

[alane1994]

\(\dfrac{d\mu}{dx}=3\mu\)
\(\dfrac{1}{\mu}\dfrac{d\mu}{dx}=3\)
\(\dfrac{1}{\mu}d\mu=3dx\)
Then solve yeah?

 

[Opalg]

\(\dfrac{d\mu}{dx}=3\mu\)
\(\dfrac{1}{\mu}\dfrac{d\mu}{dx}=3\)
\(\dfrac{1}{\mu}d\mu=3dx\)
Then solve yeah?

Yes! (Smile)

 

[alane1994]

\(C=(3x^2y+y^3)e^{3x}\)

EDIT: Is this correct?