Tricky integration, possibly Integration Factor

**craig** · May 23rd 2009, 08:12 AM

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

**Moo** · May 23rd 2009, 08:18 AM

Hello,

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 ?

**craig** · May 23rd 2009, 08:23 AM

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 ?

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

**Moo** · May 23rd 2009, 08:27 AM

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

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

)

**craig** · May 25th 2009, 06:30 AM

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

**Jester** · May 25th 2009, 06:45 AM

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 $\int\frac{dy}{4+y^2} = \frac{1}{2} \tan^{-1}\frac{y}{2} + c$

**craig** · May 25th 2009, 06:58 AM

*hits head on desk*

Of course it is, it's even in my formula book!

Thank for pointing that out, I think I'll wake myself up before attempting any more of these

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