Answer to Question #255119 in Differential Equations for sach

Question

Find the series solution of the differential equation

x²(𝑥+2)𝑦′′−𝑥𝑦′+(1+𝑥)𝑦=0.

Answer 1

Given

$x^{2}(x+2)y''-xy'+(1+x)y=0$ this is equation (i)

Let $y=\displaystyle\sum_{n=0}^{\infty}C_nx^n$

Then $y'=\displaystyle\sum_{n=1}^{\infty}C_nnx^{n-1}$

And $y''=\displaystyle\sum_{n=2}^{\infty}C_nn(n-1)x^{n-2}$

Substituting the above into our equation (i)

$x^{2}(x+1)\displaystyle\sum_{n=2}^{\infty}C_nn(n-1)x^{n-2}-x\displaystyle\sum_{n=1}^{\infty}C_nnx^{n-1}$

$+(1+x)\displaystyle\sum_{x=0}^{\infty}C_nx^n=0$

$\displaystyle\sum_{n=2}^{\infty}C_nn(n-1)x^{n+1}-\displaystyle\sum_{n=1}^{\infty}C_nnx^{n}+\displaystyle\sum_{n=0}^{\infty}C_nx^n+\displaystyle\sum_{n=0}^{\infty}C_nx^{n+1}=0$

$\displaystyle\sum_{n=3}^{\infty}C_{n-1}(n-1)(n-2)x^{n}+\displaystyle\sum_{n=1}^{\infty}C_nnx^{n}+\displaystyle\sum_{n=0}^{\infty}C_nx^n$

$+\displaystyle\sum_{n=1}^{\infty}C_{n-1}x^{n}=0$

$\displaystyle\sum_{n=3}^{\infty}C_{n-1}(n-1)(n-2)x^{n}-C_1x-2C_2x^{2}-\displaystyle\sum_{n=3}^{\infty}C_nnx^{n}+C_0$

$+C_1x+C_2x^{2}+\displaystyle\sum_{n=3}^{\infty}C_nx^n+C_0x+C_1x^{2}+\displaystyle\sum_{n=3}^{\infty}C_{n-1}x^n=0$

$C_0+C_0x+(-2C_2+C_2+C_1)x^{2}+[\displaystyle\sum_{n=3}^{\infty}(n-1)(n-2)C_{n-1}$

$-nC_n+C_nx^n+C_{n-1}]x^n=0$

Equating each coefficient by 0

$C_0=0$

$-2C_2+C_2+C_1=0$

$-C_2+C_1=0$

$C_1=C_2$

For $n>2$ ,

$(n-1)(n-2)C_{n-1}-nC_n+C_n+C_{n-1}=0$

$[(n-1)(n-2)+1]C_{n-1}+(1-n)C_n=0$

$C_n=\frac{(n-1)(n-2)+1}{(n-1)}C_{n-1}$

$C_n=[(n-2)+\frac{1}{(n-1)}]C_{n-1}$

$C_3=(1+\frac{1}{2})C_2$

$C_3=\frac{3}{2}C_2$

$C_4=(2+\frac{1}{3})C_3$

$C_4=\frac{7}{3}C_3$

$C_5=(3+\frac{1}{4})C_4$

$C_5=\frac{13}{4}C_4$

$C_6=(4+\frac{1}{5})C_5$

$C_6=\frac{21}{5}C_5$

Therefore, the series solution of the differential equation is

$y=\frac{3}{2}C_2x^3+\frac{7}{3}C_3x^4+\frac{13}{4}C_4x^5+\frac{21}{5}C_5x^6+...$