How To Integrate This?

• basty
In summary,The author suggests integrating by parts to solve an equation in standard form. Once the equation is in the standard form, the integrating factor is e^{\int (-3) dx} = e^{\int x^{-4}}. This allows for the solution to be found by multiplying the equation by the integrating factor and solving for the variable.f

Homework Statement

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

?

Explain please and give me some hints.

The Attempt at a Solution

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?

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Homework Statement

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

?

Explain please and give me some hints.

The Attempt at a Solution

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?

jedishrfu
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.

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.

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

So show us your work so far

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.

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.

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...

So this is for a differential equations course?

I rewrite the example from my differential equations book into this thread, below.
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.

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.##
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?

jedishrfu
Thanks Mark44.

This is the another example from the book:
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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Another example
Solve
##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.##
Again it shows that

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

or

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

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

I don't understand. Please explain more.

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.##
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.

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

Similar thing in another examples he provides in the book.
If d/dx(?) = 0, what does ? have to be?
I don't understand. Please explain more.
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.

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

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

I don't understand. Please explain more.

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 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.