# Tricky integration, possibly Integration Factor

• May 23rd 2009, 08:12 AM
craig
Tricky integration, possibly Integration Factor
Hi, for this problem I would normally use the Integration factor, but as it's $y^2$ I'm not too sure what to do.

$x\frac{dy}{dx} - y^2 = 4$

Have divided through by x, as I would normally in a question like this:

$\frac{dy}{dx} - \frac{y^2}{x} = \frac{4}{x}$.

Not too sure where to go from here however...

Thanks in advance for the help :)
• May 23rd 2009, 08:18 AM
Moo
Hello,
Quote:

Originally Posted by craig
Hi, for this problem I would normally use the Integration factor, but as it's $y^2$ I'm not too sure what to do.

$x\frac{dy}{dx} - y^2 = 4$

Have divided through by x, as I would normally in a question like this:

$\frac{dy}{dx} - \frac{y^2}{x} = \frac{4}{x}$.

Not too sure where to go from here however...

Thanks in advance for the help :)

Here is how you can do :)

$x\frac{dy}{dx}=4+y^2$

$\Rightarrow \frac{dy}{4+y^2}=\frac{dx}{x}$

Now, can you deal with that ? :p
• May 23rd 2009, 08:23 AM
craig
Quote:

Originally Posted by Moo
Hello,

Here is how you can do :)

$x\frac{dy}{dx}=4+y^2$

$\Rightarrow \frac{dy}{4+y^2}=\frac{dx}{x}$

Now, can you deal with that ? :p

Thanks for the reply :) I'm afraid I'm not familiar with that notation though.

Is that the same as writing

$\int \frac{1}{4+y^2} = \int \frac{1}{x}$ ?

Sorry for the lack of understanding :p
• May 23rd 2009, 08:27 AM
Moo
Quote:

Originally Posted by craig
Thanks for the reply :) I'm afraid I'm not familiar with that notation though.

Is that the same as writing

$\int \frac{1}{4+y^2} = \int \frac{1}{x}$ ?

Sorry for the lack of understanding :p

Yes, but you have to write the dy and dx !

From the equality I wrote, you can integrate from 0 to t :

$\int_0^t \frac{1}{4+y^2} ~dy=\int_0^t \frac 1x ~dx$

This is called a separable ODE (see here : http://www.usna.edu/MathDept/CDP/DE/...Separation.pdf :))
• May 25th 2009, 06:30 AM
craig
Quote:

Originally Posted by Moo
Yes, but you have to write the dy and dx !

From the equality I wrote, you can integrate from 0 to t :

$\int_0^t \frac{1}{|4+y^2|} ~dy=\int_0^t \frac 1x ~dx$

This is called a separable ODE (see here : http://www.usna.edu/MathDept/CDP/DE/...Separation.pdf :))

Thank you for this :)

So for the indefinite integral $\int \frac{1}{4+y^2} ~dy=\int \frac 1x ~dx$

Would you get?

$\frac{1}{2y} \ln{(y^2 + 4)} = \ln(x) + C$ ?

Thanks again
• May 25th 2009, 06:45 AM
Jester
Quote:

Originally Posted by craig
Thank you for this :)

So for the indefinite integral $\int \frac{1}{4+y^2} ~dy=\int \frac 1x ~dx$

Would you get?

$\frac{1}{2y} \ln{(y^2 + 4)} = \ln(x) + C$ ?

Thanks again

No $\int\frac{dy}{4+y^2} = \frac{1}{2} \tan^{-1}\frac{y}{2} + c$
• May 25th 2009, 06:58 AM
craig