Question

differential equations linear equations
dy/dx+2xy=x^3
reviewing for test tomorrow. I have problems with two questions of my review please help!

Answer 1

\[dy/dx+2xy=x^3\]
integrating factor is \[e^{(x^2)}\]
so...

Answer 2

\[e^{(x^2)}y=\int\limits e^{(x^2)}x^3dx+c\]

Answer 3

Wait wait wait a minute, where did you get \(e^{x^2}x^3\)?

Answer 4

linear integrating factor is \(\large e^{\int2x\mathrm dx}=e^{x^2}\)

Answer 5

Oh yes yes yes, I see.

Answer 6

so to evaluate that integral you might want to substitute u=x^2

Answer 7

yes i tried u sub with u=x^2 but then du=2xdx and we dont have that

Answer 8

Rearrange things, maybe it'll jump out at you :)
\[\Large \int\limits x^2\cdot e^{x^2}(x\;dx)\]

Answer 9

So the exponent is u.
Is there another u somewhere perhaps? :D

Answer 10

My brain is either warped or damn u gaiz r good :)

Answer 11

Dat's ok abs +_+ You get prettier every day

Answer 12

lolz

Answer 13

:')
I knew i made the right choice when I gave out that medal

Answer 14

ok so u=x^2 du=2xdx 1/2du=xdx
so integral becomes..
\[1/2\int\limits u e^{(x^2)}du\]??

Answer 15

The exponent becomes u also right? :D

Answer 16

oh yeah

Answer 17

\[\Large \frac{1}{2}\int\limits u e^{u}\;du\]
Mmm good good looks like you're on the right track.
That should make integration by parts a bit easier.

Answer 18

so after tabular its 1/2ue^u-e^u

Answer 19

Looks good, don't forget to distribute the 1/2.

Answer 20

\[x^2e^{(x^2)}/2-e^{(x^2)}/2\]

Answer 21

Good good.
We don't want to forget the + C before we do our next step.

Answer 22

\[y=x^2/2-1/2+c\]

Answer 23

Woops!
That's why we wanted to add the +C before dividing through! XD
Don't forget to divide the C.

We can't absorb a variable term into the C.

Answer 24

oh yeah so ce^(-x^2)

Answer 25

awesome thank you zepdrix now i only have one more problem then i can call it a night!

Answer 26

cool \c:/

Answer 27

im about to open my last question if you would like to help real fast....i would appreciate it!