Respuesta :
We're given
[tex]\displaystyle y_1(x) = 1 + \frac x3 + \frac{x^2}{24} + \frac{x^3}{360} + \cdots = \sum_{n=0}^\infty a_nx^n[/tex]
so let's see if we can find a closed form for the n-th term's coefficient.
Notice that
[tex]\displaystyle a_0 = 1 \\\\ a_1 = \frac13 = \frac1{1\times3} \\\\ a_2 = \frac1{24} = \frac1{(1\times3) \times (2\times4)} \\\\ a_3 = \frac1{360} = \frac1{(1\times3) \times (2\times4) \times (3\times5)}[/tex]
If the pattern continues, the next few terms are likely
[tex]\displaystyle a_4 = \frac1{8640} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6)} \\\\ a_5 = \frac1{302400} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7)} \\\\ a_6 = \frac1{14515200} = \frac1{(1\times3) \times (2\times4) \times (3\times5) \times (4\times6) \times (5\times7) \times (6\times8)}[/tex]
which leads up to the n-th term,
[tex]\displaystyle a_n = \frac1{(1\times3) \times (2\times4) \times \cdots \times (n\times(n+2))} = \frac2{n!(n+2)!}[/tex]
where the numerator is multiplied by 2 in order to "complete" the factorial pattern in (n + 2)!.
So we have
[tex]\displaystyle y_1(x) = \sum_{n=0}^\infty \frac2{n!(n+2)!} x^n[/tex]
Now we use reduction of order to find a linearly independent solution of the form [tex]y_2(x) = v(x)y_1(x)[/tex], with derivatives
[tex]\displaystyle \frac{\mathrm dy_2}{\mathrm dx} = v(x) \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm dv}{\mathrm dx} \\\\ \frac{\mathrm d^2y_2}{\mathrm dx^2} = v(x) \frac{\mathrm d^2y_1}{\mathrm dx} + 2 \frac{\mathrm dv}{\mathrm dx} \frac{\mathrm dy_1}{\mathrm dx} + y_1(x) \frac{\mathrm d^2v}{\mathrm dx^2}[/tex]
Substitute [tex]y_2[/tex] and its derivatives into the DE, and simplify the resulting expression to get a DE in terms of v(x) :
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} + \left(x\frac{\mathrm d^2y_1}{\mathrm dx^2}+3\frac{\mathrm dy_1}{\mathrm dx}-y_1\right)v = 0[/tex]
but since we know [tex]y_1(x)[/tex] satisfies the original DE, the last term vanishes and we're left with
[tex]\displaystyle x y_1 \frac{\mathrm d^2v}{\mathrm dx^2} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)\frac{\mathrm dv}{\mathrm dx} = 0[/tex]
Reduce the order by substituting [tex]w(x)=\dfrac{\mathrm dv}{\mathrm dx}[/tex] to get yet another DE in w(x) :
[tex]\displaystyle x y_1 \frac{\mathrm dw}{\mathrm dx} + \left(2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1\right)w = 0[/tex]
This equation is separable:
[tex]\displaystyle \frac{\mathrm dw}w = - \frac{2x\frac{\mathrm dy_1}{\mathrm dx}+3y_1}{xy_1}\,\mathrm dx \\\\ \frac{\mathrm dw}w = -\left(\frac2{y_1}\frac{\mathrm dy_1}{\mathrm dx} + \frac3x\right)\,\mathrm dx[/tex]
From here you would integrate to solve for w(x), then integrate again to solve for v(x), and finally for [tex]y_2(x)[/tex] by multiplying [tex]y_1(x)[/tex] by v(x). Using the fundamental theorem of calculus, you would find
[tex]\displaystyle \ln|w| = -2 \int_1^x \frac{{y_1}'(\xi)}{y_1(\xi)}\,\mathrm d\xi - 3\ln|x| + C_1 \\\\ w = \frac{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[/tex]
so that you end up with
[tex]\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)[/tex]
But the second term is already accounted for by [tex]y_1(x)[/tex] itself, so the second solution is
[tex]\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}[/tex]
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.
