Question

In Problems, 7-18 find two power series solutions of the given differential equation about the ordinary point $\left(x^2+2\right)y^{\text{'}\text{'}}+3x{y}^{\text{'}}-y=0$

Short answer

Therefore, the solution is:

$$\begin{gathered} y_1\left(x\right)=c_0\left[1+\frac{1}{4}{x}^2-\frac{7}{4·4!}{x}^4+\frac{23·7}{8·6!}{x}^6-\cdots \right] \\ {y}_2\left(x\right)=c_1\left[x-\frac{1}{6}{x}^3+\frac{14}{2·5!}{x}^5-\frac{34·14}{4·7!}{x}^7-\cdots \right] \end{gathered}$$

Step-by-step solution

Given Information

The given differential equation is;

$\left(x^2+2\right)y\text{'}\text{'}+3xy\text{'}-y=0$

Plug into the differential equation

$y\left(x\right)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+\cdots =\sum _{n=0}^{\infty }{c}_n{x}^n\text{(*)}$

Differentiating $\text{(*)}$and translating the index by one and two respectively, we obtain

$y\text{'}=\sum _{n=1}^{\infty }n{c}_n{x}^{n-1} \dots \left(1\right)$

$y\text{'}\text{'}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2} \dots \left(2\right)$$y\text{'}\text{'}=\sum _{n=2}^{\infty }n(n-1)c_n{x}^{n-2} \dots \left(2\right)$

Replacing the(1) and equation (2) in the given differential equation, we obtain

Identity Property

$\begin{array}{rcl}4c_2-c_0& =& 0\\ c_2& =& \frac{c_0}{4}\\ 12c_3+2c_1& =& 0\\ c_3& =& -\frac{c_1}{6}\\ & & \end{array}$

localid="1655107049106" width="430">k(k-1)ck+2(k+2)(k+1)ck+2+3kck-ck=0ck+2=ck-k2-2k+12(k+2)(k+1)

Continuing the same procedure and replacing inlocalid="1655107090240" $\left(*\right),$we obtain

localid="1655107056593" width="346">y(x)=c0+c1x+14c0x2-16c1x3-74·4!c0=c01+14x2-74·4!x4+23·78·6!x6-⋯+c1x-16x3+142·5!x5-34·144·7!x7-⋯x4+142·5!c1x5+23·78·6!c1x6-34·144·7!c1x7+⋯

Therefore, the solution is:

localid="1655107041835" width="318">y1(x)=c01+14x2-74·4!x4+23·78·6!x6-⋯y2(x)=c1x-16x3+142·5!x5-34·144·7!x7-⋯