Thread: Tricky integration, possibly Integration Factor

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

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

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

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

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

Thanks in advance for the help

Hello,

Originally Posted by craig

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

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

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

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

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

Thanks in advance for the help

Here is how you can do

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

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

Now, can you deal with that ?

Originally Posted by Moo

Hello,

Here is how you can do

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

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

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

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

Sorry for the lack of understanding

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

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

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 :

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

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

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 :

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

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

Would you get?

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

Thanks again

Originally Posted by craig

Thank you for this

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

Would you get?

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

Thanks again

No $\displaystyle \int\frac{dy}{4+y^2} = \frac{1}{2} \tan^{-1}\frac{y}{2} + c$

*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