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## Homework Statement

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

????

Explain please and give me some hints.

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- Thread starter basty
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- #1

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

????

Explain please and give me some hints.

- #2

jedishrfu

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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?

And show us your work.

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?

And show us your work.

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- #3

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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?## Homework Statement

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

????

Explain please and give me some hints.

## Homework Equations

## The Attempt at a Solution

- #4

jedishrfu

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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?

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

- #5

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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?

Yes, y is a function of x.

- #6

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

- #7

jedishrfu

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So show us your work so far

- #8

jedishrfu

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

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

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So show us your work so far

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

jedishrfu

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

- #11

jedishrfu

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So this is for a differential equations course?

- #12

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Solve

##\frac{dy}{dx} - 3y = 0.##

Solution

This linear equation can be solved by separation of variables. Alternatively, since the equation is already in the standard form ##[\frac{dy}{dx} + P(x) \ y = f(x)]##, we see that ##P(x) = - 3## and so the integrating factor is

##e^{\int P(x) \ dx} = e^{\int (-3) \ dx} = e^{-3x}.##

We multiply the equation by this factor and recognize that

##e^{-3x} \frac{dy}{dx} - 3 e^{-3x} y = 0## is the same as ##\frac{d}{dx}[e^{-3x}y] = 0.##

Integrating both sides of the last equation gives ##e^{-3x}y = c.## Solving for ##y## gives us the explicit solution ##y = ce^{3x}, -∞ < x < ∞.##

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

Mark44

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This is the same type of problem you were stuck on yesterday.basty said:We multiply the equation by this factor and recognize that

##e^{-3x} \frac{dy}{dx} - 3 e^{-3x} y = 0## is the same as ##\frac{d}{dx}[e^{-3x}y] = 0.##

What they are saying is that if the derivative of <something> equals 0, then <something> must be a constant, so e

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

- #14

jedishrfu

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Thanks Mark44.

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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.##

Solve

##\frac{dy}{dx} - 3y = 6.##

Solution

The associated homogeneous equation for this DE was solved in above example. Again the equation is already in the standard form, and the integration factor is still ##e^{\int (-3) dx} = e^{-3x}.## This time multiplying the given equation by this factor gives

##e^{-3x}\frac{dy}{dx} - 3 e^{-3x}y = 6e^{-3x},## which is the same as ##\frac{d}{dx}[e^{-3x}y] = 6e^{-3x}##

Integrating both sides of the last equation gives ##e^{-3x}y = -2e^{-3x} + c## or ##y = -2 + ce^{3x}, - ∞ < x < ∞.##

Again, the above example shows

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

or

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

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- #16

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Again it shows thatSolve

##x \frac{dy}{dx} - 4y = x^6e^x.##

Solution

Dividing by ##x,## we get the standard form

##\frac{dy}{dx} - \frac{4}{x}y = x^5e^x.##

From this form we identify ##P(x) = -\frac{4}{x}## and ##f(x) = x^5e^x## and further observe that ##P## and ##f## are continuous on ##(0, ∞).## Hence the integrating factor is

##e^{-4 \int \frac{dx}{x}} = e^{-4 \ln x} = e^{\ln x^{-4}} = x^{-4}##

Now we multiply ##\frac{dy}{dx} - \frac{4}{x}y = x^5e^x## by ##x^{-4}## and rewrite

##x^{-4}\frac{dy}{dx} - 4x^{-5}y = xe^x## as ##\frac{d}{dx}[x^{-4}y] = xe^x##

It follows from integration by parts that the general solution defined on the interval ##(0, ∞)## is

##x^{-4}y = xe^x - e^x + c## or ##y = x^5e^x - x^4e^x + cx^4.##

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

or

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

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

I don't understand. Please explain more.

- #18

vela

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Actually, from the text, it's clear the author means to integrate with respect to ##x##. That is, you haveI 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.##

$$\int \frac{d}{dx}(e^{-3x}y)\,dx = \int 0\,dx.$$ What you wrote in the original post, which omitted the context, was confusing at best.

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

Similar thing in another examples he provides in the book.

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.I don't understand. Please explain more.If d/dx(???) = 0, what does ??? have to be?

- #19

jedishrfu

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

- #20

Mark44

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

I don't understand. Please explain more.

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.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.

https://www.physicsforums.com/threads/the-difference-between-log-and-ln.785912/

https://www.physicsforums.com/threads/integration-de.786515/#post-4939203

https://www.physicsforums.com/threads/trigonometry-question.785240/

https://www.physicsforums.com/threads/trigonoemtric-identity.785218/#post-4930693

https://www.physicsforums.com/threads/why-pi-2-1-and-pi-2-5.783165/#post-4919894

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