Question

Help with calculus, please? I will give a medal!
(posting question below)

Answer 1

\[\frac{ dy }{ dx }=xe^{ax^2}\]

Answer 2

Looking for anti-derivative?

Answer 3

My instructions just say to solve the differential equation, so maybe?

Answer 4

Sure sure sure :)\[\large\rm dy=x e^{ax^2}dx\]Integrating,\[\large\rm \int\limits dy=\int\limits x e^{ax^2}dx\]So we'll have to figure out how to deal with the right side,\[\large\rm y=\int\limits x e^{ax^2}dx\]

Answer 5

Okay that looks familiar

Answer 6

Do you always do that when you have a dy/dx?

Answer 7

Let's rewrite it like this:\[\large\rm y=\int\limits e^{\color{orangered}{ax^2}}\left(\color{royalblue}{x~dx}\right)\]

Answer 8

Ummm no,
if we had another y showing up in the problem,
and the problem was not `separable`,
meaning: if we couldn't move all y's and x's to opposite sides,
then we would need another technique.

But I guess if that's the only y showing up in the problem, the y',
then yes, it should always work out :o

Answer 9

Okay. Gotcha.

Answer 10

So our goal here is to make a substitution of some type,
let's say,\[\large\rm \color{orangered}{u=ax^2}\]And then we would like our du to look `something like` the blue stuff.

Answer 11

If the a was a number, I wouldn't have a problem solving this, but the a throws me off. Would it by chance be 2ax?

Answer 12

\[\large\rm du=2ax~ dx\]k looks good.

Answer 13

You mean that's right? Awesome!
Then you would just get...
\[\frac{1}{2a}\int\limits_{}^{}e^udu\]

Answer 14

Good good good.

Answer 15

K I think I got it.
\[y=\frac{1}{2}e^{ax^2}\]

Answer 16

Woah, where'd the 1/a go? :D

Answer 17

Oops! Just a sec

Answer 18

\[y=\frac{1}{2a}e^{ax^2}\]

Answer 19

\(\large\rm y(x)=\frac{1}{2a}e^{ax^2}+c\)

Cool, good job!

Answer 20

Dang it. I always forget the C.
Thank you so much for your help!

Answer 21

np