# How To Integrate This?

1. Dec 10, 2014

### basty

1. The problem statement, all variables and given/known data

$\int \frac{d}{dx} (x^2y)$

????

Explain please and give me some hints.

2. Relevant equations

3. The attempt at a solution

2. Dec 10, 2014

### Staff: Mentor

Have you learned integration by parts udv = uv - vdu?

On second thought it's not a good hint...

Look in your book for implicit integration.

So we know dy/dx = x^2 y right?

so what can you do here to put y on one side and x on the other?

Last edited: Dec 10, 2014
3. Dec 10, 2014

### LCKurtz

You are supposed to show an attempt at solving. Also you haven't told us several things. What variable is the integration with respect to? Is $y$ a function of $x$ or an independent variable?

4. Dec 10, 2014

### Staff: Mentor

Thanks. Good point, I made an assumption in my hint that y is a function of x.

5. Dec 10, 2014

### basty

Yes, y is a function of x.

6. Dec 10, 2014

### basty

Is $\int \frac{d}{dx} (x^2y) = x^2y \ ?$

7. Dec 10, 2014

### Staff: Mentor

So show us your work so far

8. Dec 10, 2014

### Staff: Mentor

No, that's not it. It would be if y weren't a function of x.

However y is a function of x so you have to do something more complicated.

9. Dec 10, 2014

### basty

I read this kind of integration in my differential equations book and he (the author) also does not show in detail how he got the result. He just write that

$\int \frac{d}{dx}(anything) = anything$

without showing the work of it.

10. Dec 10, 2014

### Staff: Mentor

Okay what you quoted is the fundamental theorem of calculus. Hmm I have to think about it now...

You may be right. I took this so long ago I just can't remember...

11. Dec 10, 2014

### Staff: Mentor

So this is for a differential equations course?

12. Dec 10, 2014

### basty

I rewrite the example from my differential equations book into this thread, below.
When he says "integrating both sides of the last equation..." mean $\int \frac{d}{dx}[e^{-3x}y] = \int 0.$

And the result is $e^{-3x}y = c.$

It means that $\int \frac{d}{dx}[e^{-3x}y] = e^{-3x}y$

or

$\int \frac{d}{dx}(whatever) = whatever$

Similar thing in another examples he provides in the book.

13. Dec 10, 2014

### Staff: Mentor

This is the same type of problem you were stuck on yesterday.

What they are saying is that if the derivative of <something> equals 0, then <something> must be a constant, so e-3xy = C. That's all it is.

If d/dx(???) = 0, what does ??? have to be?

14. Dec 10, 2014

### Staff: Mentor

Thanks Mark44.

15. Dec 11, 2014

### basty

This is the another example from the book:
Again, the above example shows

$\int \frac{d}{dx}[e^{-3x}y] = e^{-3x}y$

or

$\int \frac{d}{dx}[whatever] = whatever.$

Last edited: Dec 11, 2014
16. Dec 11, 2014

### basty

Another example
Again it shows that

$\int \frac{d}{dx}[x^{-4}y] = x^{-4}y$

or

$\int \frac{d}{dx}[whatever] = whatever.$

17. Dec 11, 2014

### basty

I don't understand. Please explain more.

18. Dec 11, 2014

### vela

Staff Emeritus
Actually, from the text, it's clear the author means to integrate with respect to $x$. That is, you have
$$\int \frac{d}{dx}(e^{-3x}y)\,dx = \int 0\,dx.$$ What you wrote in the original post, which omitted the context, was confusing at best.

These results are from beginning calculus. It seems a little strange that you're taking differential equations if you're not familiar with these facts already.

19. Dec 11, 2014

### Staff: Mentor

Thanks, Vela. Now, I see where I went wrong with my advice.

20. Dec 11, 2014

### Staff: Mentor

I agree with vela. You have started a number of threads with questions about differential equations, but have run into roadblocks in concepts from algebra, trig, and basic calculus. You will not be successful in studying differential equations if you don't understand the basic concepts that you should already be competent in.