Question

Given the solution
`y_(1)(x)` from EDO below, develop a second solution.


`x(d^(2)y )/(dx^(2) ) +3(dy)/(dx) -y=0,\\y_(1) (x)=1+(x)/(3) +(x^(2) )/(24) +(x^(3) )/(360) + ...`

Answer 1

We're given

`\displaystyle y_1(x) = 1 + \frac x3 + (x^2)/(24) + (x^3)/(360) + \cdots = \sum_(n=0)^\infty a_nx^n`

so let's see if we can find a closed form for the n-th term's coefficient.

Notice that

`\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1*3} \\\\ a_2 = \frac1{24} = \frac1{(1*3) * (2*4)} \\\\ a_3 = \frac1{360} = \frac1{(1*3) * (2*4) * (3*5)}`

If the pattern continues, the next few terms are likely

`\displaystyle a_4 = \frac1{8640} = \frac1{(1*3) * (2*4) * (3*5) * (4*6)} \\\\ a_5 = \frac1{302400} = \frac1{(1*3) * (2*4) * (3*5) * (4*6) * (5*7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1*3) * (2*4) * (3*5) * (4*6) * (5*7) * (6*8)}`

which leads up to the n-th term,

`\displaystyle a_n = \frac1{(1*3) * (2*4) * \cdots * (n*(n+2))} = \frac2{n!(n+2)!}`

where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.

So we have

`\displaystyle y_1(x) = \sum_(n=0)^\infty \frac2{n!(n+2)!} x^n`

Now we use reduction of order to find a linearly independent solution of the form
`y_2(x) = v(x)y_1(x)`, with derivatives

`\displaystyle (\mathrm dy_2)/(\mathrm dx) = v(x) (\mathrm dy_1)/(\mathrm dx) + y_1(x) (\mathrm dv)/(\mathrm dx) \\\\ (\mathrm d^2y_2)/(\mathrm dx^2) = v(x) (\mathrm d^2y_1)/(\mathrm dx) + 2 (\mathrm dv)/(\mathrm dx) (\mathrm dy_1)/(\mathrm dx) + y_1(x) (\mathrm d^2v)/(\mathrm dx^2)`

Substitute
`y_2` and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :

`\displaystyle x y_1 (\mathrm d^2v)/(\mathrm dx^2) + \left(2x(\mathrm dy_1)/(\mathrm dx)+3y_1\right)(\mathrm dv)/(\mathrm dx) + \left(x(\mathrm d^2y_1)/(\mathrm dx^2)+3(\mathrm dy_1)/(\mathrm dx)-y_1\right)v = 0`

but since we know
`y_1(x)` satisfies the original DE, the last term vanishes and we're left with

`\displaystyle x y_1 (\mathrm d^2v)/(\mathrm dx^2) + \left(2x(\mathrm dy_1)/(\mathrm dx)+3y_1\right)(\mathrm dv)/(\mathrm dx) = 0`

Reduce the order by substituting
`w(x)=(\mathrm dv)/(\mathrm dx)` to get yet another DE in w(x) :

`\displaystyle x y_1 (\mathrm dw)/(\mathrm dx) + \left(2x(\mathrm dy_1)/(\mathrm dx)+3y_1\right)w = 0`

This equation is separable:

`\displaystyle \frac{\mathrm dw}w = - (2x(\mathrm dy_1)/(\mathrm dx)+3y_1)/(xy_1)\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}(\mathrm dy_1)/(\mathrm dx) + \frac3x\right)\,\mathrm dx`

From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for
`y_2(x)` by multiplying
`y_1(x)` by v(x). Using the fundamental theorem of calculus, you would find

`\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = (C_1)/(x^3) \exp\left(-2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)} \,\mathrm d\xi\right)\right) \\\\ v = C_1 \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2`

so that you end up with

`\displaystyle y_2(x) = C_1 y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega + C_2y_1(x)`

But the second term is already accounted for by
`y_1(x)` itself, so the second solution is

`\displaystyle y_2(x) = \boxed{y_1(x) \int_1^x \frac1{\omega^3} \exp\left(-2 \int_1^\omega \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi\right) \,\mathrm d\omega}`

You could go the extra mile and try to find a power series expression for this solution, but that's a lot of work for little payoff IMO.