Question

Solve the equation, find an explicit solution if possible. Equation: x(dy/dx)-x^3e^(2x)-2y=0

Answer 1

If you don't know how to approach this problem, watch this great lecture. It will explain to you exactly how to solve this sort of problem.

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 2

\[\large xy'-x^3e^{2x}-2y=0\]Move some stuff around,\[\large xy'-2y=x^3e^{2x}\]Divide both sides by x to get it off of the leading y term.\[\large y'+\left(-\frac{2}{x}\right)y=x^2e^{2x}\]

Answer 3

From here we need to find an integrating factor:\[\large \mu=e^{-\int\limits\frac{2}{x}dx}\]

Answer 4

Stop there please. It is important that the questioner have the chance to take the next steps themselves.

Answer 5

ok so e=1/x^2 right?

Answer 6

sorry, μ=e^(ln(x^2))

Answer 7

then μ=1/x^2 and then you multiply it on both sides, right?

Answer 8

Yes

Answer 9

you get \[(y'/x^2)-(2y/x^3)=e^(2x)\]

Answer 10

then i'm lost...

Answer 11

Hmm looks good so far! :)
The point of introducing an `integrating factor` is that it SHOULD turn our left side into something that looks like the product rule.

We're basically trying to apply the product rule in reverse.
We have something like this:\[\large u'v+uv'\] and we want to try it like this,\[\large (uv)'\]

Answer 12

want to write it like** blah typo :3

Answer 13

What's the whole point of the integrating factor? ... right, what zepdrix wrote.

I strongly recommend you watch the lecture where this is carefully and nicely explained.

Answer 14

If you correctly found your integrating factor \(\mu\),
then the left side should simplify down to,\[\large \left(\mu \cdot y\right)'\]It's a good idea to apply the product rule from here, to make sure it gives you the two terms you had. That will let you know if you made a mistake! :)

Answer 15

ok I'm going to substitute -2/x^3 with d/dx (1/x^2) which will look like \[y* d/dx (1/x^2)+(dy/dx)/x^2=e^(2x)\]

Answer 16

I get d/dx (y/x^2)=e^(2x) which i integrate both sides and get y/(x^2) = e^(2x)/2+\[c_{1}\]

Answer 17

yes

Answer 18

and then i can divide both sides by 1/x^2

Answer 19

woot thanks!

Answer 20

yay good job c: