Question

Note that x=0 is an ordinary point of the differential equation$y"+x^{2}y+2xy=5-2x+10x^3$. Use the assumptionrole="math" localid="1664966249373" width="88" height="54">y=∑n=0xcnxnto find the general solution$y=y_c+y_p$that consists of three power series centered at x=0.

Short answer

$$\begin{gathered} y=c_0\left(1-\frac{1}{3}{x}^3+\frac{1}{3^2-2!}{x}^6-\frac{1}{3^3-3!}{x}^9+\frac{1}{3^4-4!}{x}^{12}-...\right) \\ +c_1\left(x-\frac{1}{4}{x}^4+\frac{1}{4.7}{x}^7-\frac{1}{4.7-10}{x}^{10}+\frac{1}{4.7-10-13}{x}^{13}-...\right) \\ +\left(\frac{5}{2}{x}^2-\frac{1}{3}{x}^3+\frac{1}{3^2-2!}{x}^6-\frac{1}{3^3-3!}{x}^9+\frac{1}{3^4-4!}{x}^{12}-...\right) \end{gathered}$$

Step-by-step solution

Definition of Power series solutions

If $x=x_0$is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at$x_0$, that is,

$y=\sum _{n=0}^x{C}_n{(x-x_0)}^n$

A power series solution converges at least on some interval defined by$\left|\begin{array}{c}x-x_0\end{array}\right|<R$, where R is the distance from$x_0$ to the closest singular point.

Using Maclaurin series to solve the problem

We were given the next equation

$y"+x^{2}y+2xy-5-2x=10x^3$

$x=0$is an ordinary point of the DE (1). By Theorem 6.2.1, we can find two linearly independent solutions in the next form

$y=\sum _{n=0}^{\infty }{C}_n{x}^n$

The derivations of (2) are

role="math" localid="1664967775447" $y=\sum _{n=1}^{\infty }{C}_{n}n{x}^{n-1},y=\sum _{n=2}^{\infty }{C}_{n}n(n-1)x^{n-2}$

Substitute this into (1) to obtain

$5-2x+10x^3=\sum _{n=2}^{\infty }{c}_{n}n(n-1)X^{n-2}+\underset{n=1}{\overset{\infty }{x^2\sum }}{c}_{n}n{x}^{n-1}+2\sum _{n=0}^{\infty }{c}_n{X}^n$

role="math" localid="1664969843269" $=\sum _{n=2}^{\infty }{c}_{n}n(n-1)X^{n-2}+\sum _{n=1}^{\infty }{c}_{n}n{x}^{n+1}+2\sum _{n=0}^{\infty }{c}_n{X}^{n+1}$

We can rewrite (4) changing index of the first sum inand the index of the other two sums in:

$5-2x+10x^3=\sum _{k=0}^{\infty }{c}_{k+2}(k+2)(k+1)X^k+\sum _{k=2}^{\infty }{c}_{k-1}(k-1)X^k+2\sum _{k=1}^{\infty }{c}_{k-1}{X}^k$

role="math" localid="1664969167598" $2c_2+(6c_3+2c_0)+\sum _{k=2}^{\infty }\left[\left(k+2\right)\left(k+1\right)c_{k+2}+(k+1)c_{k-1}\right]X^k$

Substitute the values

From this, we obtain the next system.

$$\begin{gathered} 2c_2=5 \\ 6c_3+2c_0=-2 \\ 12c_4+3c_1=0 \\ 20c_5+4c_2=10 \\ (k+2)(k+1)C_{k+2}+(k+1)C_{k1}=0,k=4,5,6,.... \\ Thisisequivalentto \\ \left\{C_2=\frac{5}{2} \\ {C}_3=-\frac{1}{3}{C}_0-\frac{1}{3} \\ {C}_4=-\frac{1}{4}{C}_1 \\ {C}_5=\frac{1}{2}-\frac{1}{5}{C}_2=\frac{1}{2}-\frac{1}{5},\frac{5}{2}=0 \\ {C}_{k-2}=-\frac{1}{k+2}{C}_{k-1},k=4,5,6,........\begin{array}{}\end{array}\right. \\ Fromthis,weobtain: \\ {C}_6=-\frac{1}{6}{C}_3=\frac{1}{3.6}(C_0+1)=\frac{1}{3^2.2!}{C}_0+\frac{1}{3^2.2!}{C}_0+\frac{1}{3^2.2!} \\ {C}_7=-\frac{1}{7}{C}_4=\frac{1}{4.7}{C}_1 \\ {C}_8=C_{11}=C_{14}=....=0 \\ {C}_9=-\frac{1}{9}{C}_6=\frac{1}{3^3.3!}{C}_0+\frac{1}{3^3.3!} \\ {C}_{10}=-\frac{1}{10}{C}_7=-\frac{1}{4.7.10}{C}_1 \\ {C}_{12}=-\frac{1}{12}{C}_9=\frac{1}{3^4.4!}{C}_0+\frac{1}{3^4.4!} \\ {C}_{13}=-\frac{1}{13}{C}_{10}=-\frac{1}{4.7.10.13}{C}_1 \end{gathered}$$

Therefore, the general solution is

$y=c_0\left(1-\frac{1}{3}{x}^3+\frac{1}{3^2.2!}{x}^6-\frac{1}{3^3.3!}{x}^9+\frac{1}{3^4.4!}{x}^{12}-...\right)+c_1\left(x-\frac{1}{4}{x}^4+\frac{1}{4.7}{x}^7-\frac{1}{4.7.10}{x}^{13}-...\right)+\left(\frac{5}{2}{x}^2-\frac{1}{3}{x}^3=\frac{1}{3^2.2!}{x}^6-\frac{1}{3^3.3!}{x}^9+\frac{1}{3^4.4!}{x}^{12-....}\right)$