Question

Help me solve this differential equation with Integrating Factor Technique :::
y'+2y/x=(e^(-2))/x
I need not only the result (Mathematica does that well) but the solving steps.
Thanks

Answer 1

Use WolframAlpha, it shows the steps.

Answer 2

\[dy/dx+2y/x=e ^{-2}/x\]multiply both sides by x:\[xdy/dx+2y=e ^{-2}\]\[xdy/dx=e ^{-2}-2y\]\[dy/(e ^{-2}-2y)=dx/x\]integrate both sides:\[\int dy/(e ^{-2}-2y)=\int dx/x\]\[-0.5\ln (e ^{-2}-2y)=\ln (x)+C_1\]\[\ln (e ^{-2}-2y)=-2\ln (x)+C_2\]\[\ln (e ^{-2}-2y)=\ln (x ^{-2})+C_2=\ln (x ^{-2})+\ln (C_3)=\ln(C_4x ^{-2})\]now both sides have logs, so we can remove those:\[e ^{-2}-2y=C_4x ^{-2}\]\[y=(e ^{-2}-C_4x ^{-2})/2=1/2e ^{2}+C_5/x ^{2}\]

Answer 3

wouldnt the integrating factor be:
\[e^{\int p(x) dx}\]
where p(x) is a portion of the "y" part

Answer 4

Yes.

This is straight-forward equation if you understand the method. I strongly suggest you review the method, from your lecture notes, text book, or this good lecture:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 5

Misspelling in equation. It should be :::
y'+2y/x=(e^(-x^2))/x

Answer 6

No problem. When you've found the integrating factor this will be quite straightforward. But you've got to understand the method. I strongly recommend watching the lecture I posted. It's good stuff.

Answer 7

Still, it can't get full solution, here is why:
1) y'(x)+(2 y(x))/x = e^(-x^2)/x
2) Integrating factor = C*x, we take C=1 int.fact.=x=a ("a" so it would be easier to write next equations)
3) a*y'(x)+a*(2 y(x))/x = a*e^(-x^2)/x
4) We need a*[y'+(2y/x]=(ay)'
5) from above we get (ay)'=a*e^(-x^2)/x
6) y=(1/a) *[ int (a * e^(-x^2)/x * dx) + C]
7) inserting a=x
y=C/x + 1/x * [int (x * {(e^(-x^2))/x} dx]

and, the integer is some kind of error function.

Answer 8

No, the integrating factor is not Cx

Answer 9

True that! :D
it's Cx^2

Answer 10

In the notation of the lecture, p(x) = 2/x, hence the integrating factor is x^2

Answer 11

Yes, but you don't need the constant. x^2 is just fine.

Answer 12

It's all uphill from here. Thanks.

Answer 13

jamesj - thanks for pointing out the lecture - I just listened to it and now understand this method.
we have a general first order linear equation of the form:\[y\prime+p(x)y=q(x)\]an we need to find an integrating factor u(x) such that u'(x)=p(x)u(x). this means:\[u(x)=e^{\int p(x)dx}\]and the original equation can be written as:\[(u(x)y)\prime=q(x)\]in this question we have:\[p(x)=2/x\]\[q(x)=e^{-x^{2}}/x\]so\[u(x)=e^{\int p(x)dx}=e^{\int 2/xdx}=e^{2\ln x}=e^{\ln x^{2}}=x^{2}\]so we have:\[(x^{2}y)\prime=e^{-x^{2}}/x\]integrate both sides to get:\[x^{2}y=-1/2e^{x^{2}}+C\]giving us the final solution:\[y=C/x^{2}-1/2x^{2}e^{x^{2}}\]

Answer 14

Yes. But you made a small error at this step:
\[ (x^2y)' =xe^{−x^2} \]
and a couple of sign errors after that step as well.

Answer 15

ah yes! - thanks, it should have been:\[(x^{2}y)'=xe^{-x^{2}}\](I forgot to multiple BOTH sides by u(x)!). this then gives:\[x^{2}y=-e^{-x^{2}}/2+C\]\[y=C/x^{2}-e^{-x^{2}}/2x^{2}=C/x^{2}-1/2x^{2}e^{x^{2}}\]hopefully I got it right this time :-)

Answer 16

Close! But a sign error in the last step. It should be

\[ y = \frac{C}{x^2} - \frac{1}{2x^2} e^{-x^2} \]

Answer 17

I think I have the same answer as you - the component of my answer is supposed to be:\[-1/(2x^{2}e^{x^{2}})\]i.e. -1 over the entire terms 2x^2e^x^2

Answer 18

I see

Answer 19

thanks for the pointers - I have learnt something new today.

Answer 20

good! These sorts of ODEs are fun.