More Calculus

[SoundTheSurrender]

Differential equations

(x) (dy)/(dx) = x^(2) e^(5x)

am I allowed to do this?


/ x dx = / x^(2) e^(5x) dy ???



/ = integrate




and for this problem.

/ 3x^(2) e^(x^2)


would I use u as x^(2)

? I don't know how to continue if I do..

 

[SpecialEd]

Problem 1:

separate:

dy=xe^(5x)dx

use integration by parts on the right

not sure on the second one.

 

[TallBill]

Its been 5 years since I took Calc 2, and I dont remember a thing. I'll be taking it again soon enough though

 

[Ika]

My god, it's only been a year and 3 months since I took calculus and I've forgotten everything already! God help me!

 

[ruffilb]

Originally posted by: Aflac
My god, it's only been a year and 3 months since I took calculus and I've forgotten everything already! God help me!

For me it's only been THREE months.

God help ME!

 

[SoundTheSurrender]

Originally posted by: SpecialEd
Problem 1:

separate:

dy=xe^(5x)dx

use integration by parts on the right

not sure on the second one.

where did the x^2 go and the x? sorry I get lost

Thanks!

 

[SpecialEd]

your teacher may not have had this in mind, but for the second problem:

write e^(x^2) as the power series sum(x^(2n)/n!)

Then your problem reduces to the integral of 3*sum((x^(2n+3))/(n!))

Then integrate.

You answer will be in the form of a power series but it will converge at all real numbers.

 

[SpecialEd]

Originally posted by: djmihow

Originally posted by: SpecialEd
Problem 1:

separate:

dy=xe^(5x)dx

use integration by parts on the right

not sure on the second one.

where did the x^2 go and the x? sorry I get lost



Thanks!

i divided both sides by x, and x^2 becomes x on the right

edit: my message was mixed up

 

[AyashiKaibutsu]

edit: nm think I made a mistake

 

[AyashiKaibutsu]

Edit: nm I'm pretty sure I made a mistake : )

 

[phatj]

3x^(2) e^(x^2)

let u = x^2
du = 2x * dx

(3/2)*u * e^(u)
then IBP

 

[phatj]

actually you can just factor out the 3/2 and get

(3/2) * integral [ u * e^u

 

[SpecialEd]

Originally posted by: phatj
3x^(2) e^(x^2)

let u = x^2
du = 2x * dx

(3/2)*u * e^(u)
then IBP

that u-sub is incorrect. you need x^3 * e^(x^2) * dx inorder to get u * e^u * du

 

[SoundTheSurrender]

Thanks, I got the first one settled, I can't do the 2nd one like you first posted specialed, we never did it in that notion..

I'm confused on the x^3 ...

is that because you did x^2 + 1 / 3 so its x^3?


so u = ?

 

[SpecialEd]

Originally posted by: djmihow
Thanks, I got the first one settled, I can't do the 2nd one like you first posted specialed, we never did it in that notion..

I'm confused on the x^3 ...

is that because you did x^2 + 1 / 3 so its x^3?


so u = ?

I was just showing why phatj's solution was incorrect. If the problem you asked was the integral of x^3*e^(x^2), then what he said would work.

But your original problem of the integral of x^2*e^(x^2) is more difficult. I don't see any obivious substitutions.

 

[SoundTheSurrender]

alrighty,

what about this then..



integrate from 6 to infinite (2x-5)/(x^2-5x) dx

u = x^(2) - 5

du = 2x - 5 dx ?

 

[Elderly Newt]

Originally posted by: djmihow
alrighty,

what about this then..



integrate from 6 to infinite (2x-5)/(x^2-5x) dx

u = x^(2) - 5

du = 2x - 5 dx ?

shouldnt it be
du/dx = 2x - 5
then
du = (2x - 5) dx
dx = du/(2x - 5)

but im tired. so who knows.

 

[SoundTheSurrender]

aggg

 

[SoundTheSurrender]

are you say it should be

1 / (2x-5) = dx ?

 

[hypn0tik]

Originally posted by: djmihow
alrighty,

what about this then..



integrate from 6 to infinite (2x-5)/(x^2-5x) dx

u = x^(2) - 5

du = 2x - 5 dx ?

Use a partial fraction expansion for this one.

(2x-5)/(x^2-5x) = (2x-5)/[(x)(x-5)].

(2x-5)/[(x)(x-5)] = A/x + B/(x-5)

Solve for A and B and integration should follow easily from there.

Edit: Ugh, I'm a noob. There's a simpler way.

Let u = x^2-5x
Then, du = 2x-5 dx

Substituting transforms the integral into 1/u du.

Make sure you have the right limits after substitution.