Question

Question on differential equations solving with Power series

Answer 1

y" - 2xy' -2y = x

Answer 2

So my real question when it come to this is that I know when we have something on the other side of the equals sign other than zero....we need to turn that into a power series (like with e^x) and then solve that way....however how do we go about doing it with just x? I don't believe there is a power series for x or at least I haven't found it

Answer 3

Have you tried the math section?

Answer 4

Ahh k, well i don't know what the other person did so i guess i'll do it from scratch using an integration factor to see if i get the same solution

y'=2xy+x^3

Since this is non-seperable we will have to arrange it to the form

y' + p(x)y = q(x)
y' - 2xy = x^3

integration factor
u(x) = e ^ [ ∫ p(x) dx]
= e ^ [ -x^2]

So i get the same factor. I think it'll cancel down with q being x^3 we'll see.

Multiplying the integration factor through
u(x) [ y' + p(x)y] = u(x) * q(x)

Simplifying
(u(x)*y)' = u(x)*q(x)
u(x)*y = ∫ [u(x) * q(x) ]*dx

e ^ [ -x^2] y = ∫ [e ^ [ -x^2] * x^3 ]dx

This looks a little messy but the right hand side sets up perfectly for integration by parts.
You can do it without the substitution which i find quicker especially for exams but i guess the person who gave you the other solution used the substitution as follows

let u = - x^2

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

RHS
= ∫ [e ^ [ -x^2] * x^3 ]dx

= ∫ e ^ [ -x^2] * ( -x^2 ) * -x * dx

= ∫ e ^ [ -x^2] * ( -x^2 ) * -2x * dx * 1/2

= ∫ [e ^ [ u] * u ]du

So the x^3 became the -x^2 * -x which all cancelled down nicely

= ∫ [e ^ [ u] * u ]du

This is just integration by parts

∫ [e ^ [ u] * u ]du = e ^ [ u] * u - ∫ [e ^ [ u] * du
= e ^ [ u] * u - e ^ [ u] + C

= e ^ [ -x^2] * -x^2 - e ^ [ -x^2] + C

Then fixing it all up

LHS = RHS
e ^ [ -x^2] y = ∫ [e ^ [ -x^2] * x^3 ]dx

e ^ [ -x^2] y = e ^ [ -x^2] * -x^2 - e ^ [ -x^2] + C

y = -x^2 - 1 + C*e ^ [ x^2]

Answer 5

Very nice work @tanijah