Solve the Initial Value Problem

[TheSpaceGuy]

Homework Statement

y' = (2x) / (y+(x^2)y) y(0) = -2

The Attempt at a Solution

I tried doing this by finding the Integrating factor and I got that to be u = -1-x^2 by using the
(My - Nx) / N formula. Using this did not work out for me and I'm not seeing the other approach. Please help me out. As always, I greatly appreciate it.

 

[lineintegral1]

The equation is separable. If I understand your post, you have,

\frac{dy}{dx} = \frac{2x}{y+yx^2}

Therefore,

dy(y+yx^2)=2xdx

Factoring, anyone?

 

[TheSpaceGuy]

Wow that does make it a lot easier. I was practicing using the Integrating factor approach so I didn't even try seeing it that way, my bad. So is there a way to get an Integrating factor for this problem by making it: (2x)dx - (y + (x^2)y)dx = 0 ?

 

[lineintegral1]

Sure, if it is not exact, then you should follow the procedure for calculating the integrating factor for non-exact equations.. If it is exact (which it isn't), then you would simply find the potential function. If you need assistance in the integrating factor procedure, let me know.

This is great practice though. You should be able to solve many of these equations using all of the methods you have learned up until this point. Hopefully, all of your answers will be the same.

 

[hunt_mat]

Use:

<br /> \frac{dy}{dx} = \frac{2x}{y+yx^2}\Rightarrow y\frac{dy}{dx}=\frac{2x}{1+x^{2}}<br />

Integrate w.r.t. x

 

[TheSpaceGuy]

Thanks for the assist hunt_mat. And lineintegral1, I had a general question about finding the integrating factor for the non-exact equations. I know how to find the integrating factor when it will be dependent on x : du/dx = My - Nx / N * u where u is the integrating factor. But what about when its dependent on y? Is it as simple as just writing dy instead of dx?

 

[hunt_mat]

Not too sure what you're saying, the integrating factor method can only be applied to equations of te form:

<br /> \frac{dy}{dx}+P(x)y=Q(x)<br />

If it's anything else from that tthen you require another method. Perhaps you would like to give us an example of the sort of thing you were thinking about.

 

[pergradus]

Not too sure what you're saying, the integrating factor method can only be applied to equations of te form:

<br /> \frac{dy}{dx}+P(x)y=Q(x)<br />

If it's anything else from that tthen you require another method. Perhaps you would like to give us an example of the sort of thing you were thinking about.

Yep. The form of the equation is important - different forms require different methods. As mentioned above this is easily separable and then you simply integrate both sides.

 

[hunt_mat]

I know.

 

[TheSpaceGuy]

I was referring to finding the integrating factor of Exact equations. With the form:
(M)dx + (N)dy = 0 and then to find the integrating factor you do
du / dx = (My - Mx) / N * u where u is the integrating factor and when I write My I mean M derived with respect to y. An example would be a problem like:
(3xy+y^2) + (x^2 +xy)y' = 0. This problem has a integrating factor of u = x. I wanted to know what if a problem had the integrating factor u = y? Would I use the same procedure to get it? Thanks.

 

[TheSpaceGuy]

Okay I just looked it up in my book and there is a procedure to do that.
If (Nx-My) / M = Q where Q is a function of y only then the integrating factor of the equation is u(y) = exp (int) Q(y) dy