Solving Separable ODEs: How to Integrate with Functions of t?

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I understand how to integrate this: ∫y2dy.

I don't understand how to integrate this:
di(t)/dt = i(t)p(t)
intergrate((di(t)/dt/i(t))*dt = p(t)dt) (see this image: http://i.imgur.com/OdKI309.png)

how do you perform the intergral on the left, seeing as as it not dt, but di(t)?

thanks
 
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dgamma3 said:
I understand how to integrate this: ∫y2dy.

I don't understand how to integrate this:
di(t)/dt = i(t)p(t)
intergrate((di(t)/dt/i(t))*dt = p(t)dt) (see this image: http://i.imgur.com/OdKI309.png)

how do you perform the intergral on the left, seeing as as it not dt, but di(t)?

thanks
Can you do this integration? ##\int \frac{du}{u}?##

BTW, there are no such words in English as "intergrate" and "intergral."
 
Mark44 said:
Can you do this integration? ##\int \frac{du}{u}?##

BTW, there are no such words in English as "intergrate" and "intergral."

Yes I can do that.

But I do not understand how do integrate (what word do I use?)
∫di(t)/i(t)
 
Last edited:
dgamma3 said:
Yes I can do that.

But I do not understand how do integrate (what word do I use?)
∫di(t)/i(t)
This is essentially the same as what I wrote.
##\int \frac{du}{u}## is the same as ##\int \frac{du(t)}{u(t)}##. The only difference is that in the second integral, it is made explicit that u is a function of t.
 
Mark44 said:
This is essentially the same as what I wrote.
##\int \frac{du}{u}## is the same as ##\int \frac{du(t)}{u(t)}##. The only difference is that in the second integral, it is made explicit that u is a function of t.
Thank you Mark!
 
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