Question

In Problems 15 and 16 solve the given initial-value problem.

(X + 2)y" + 3y = 0, y(0) = 0 , y' (0) = 1

Short answer

Therefore, the solution is

$\mathrm{y}=\mathrm{x}-\frac{1}{4}{\mathrm{x}}^3+\frac{1}{16}{\mathrm{x}}^4+\dots$

Step-by-step solution

Given Information

The given value is

$(x+2)y^{\text{'}\text{'}}+3y=0,y\left(0\right)=0,y^{\text{'}}\left(0\right)=1$.

Use Differentiate

We were given the next equation

$(\mathrm{x}+2){\mathrm{y}}^{\text{'}\text{'}}+3\mathrm{y}=0,\mathrm{y}\left(0\right)=0,{\mathrm{y}}^{\text{'}}\left(0\right)=1----\left(1\right)$

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

$\mathrm{y}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}----\left(2\right)$

The derivations of (2) are

${\mathrm{y}}^{\text{'}}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{\mathrm{nx}}^{\mathrm{n}-1},{\mathrm{y}}^{\text{'}\text{'}}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}\mathrm{n}(\mathrm{n}-1){\mathrm{x}}^{\mathrm{n}-2}----\left(3\right)$

Substitute this into (1) to obtain

We can rewrite (4) changing index of the first sum in$\mathrm{k}=\mathrm{n}-1$ and index of the first sum in$\mathrm{k}=\mathrm{n}-2:$

Identity property

From this, we can see that

$4{\mathrm{c}}_2+3{\mathrm{c}}_0=0\to {\mathrm{c}}_2=\frac{3}{4}{\mathrm{c}}_0$

${\mathrm{c}}_{\mathrm{k}+2}=-\frac{1}{2(\mathrm{k}+2)}{\mathrm{c}}_{\mathrm{k}+1}-\frac{1}{2(\mathrm{k}+2)(\mathrm{k}+1)}{\mathrm{c}}_{\mathrm{k}},\mathrm{k}\in \mathrm{\mathbb{N}}----\left(6\right)$

Let $c_0=0$and ${\mathrm{c}}_1=1.$From this, we find

$c_2=0,c_3=-\frac{1}{4},c_4=\frac{1}{16},c_5=0,...$

Let ${\mathrm{c}}_0=1$and${\mathrm{c}}_1=0.$ From this, we find

${\mathrm{c}}_2=-\frac{3}{4},{\mathrm{c}}_3=\frac{1}{8},{\mathrm{c}}_4=\frac{1}{16},{\mathrm{c}}_5=-\frac{9}{320},\dots$

Therefore, the general solution is

$y=C_0\left(1-\frac{3}{4}{x}^2+\frac{1}{8}{x}^3-\frac{1}{16}{x}^4-\frac{9}{320}{x}^5+\dots \right)+C_1\left(x-\frac{1}{4}{x}^3+\frac{1}{16}{x}^4+\dots \right)----\left(7\right)$

Its derivation is

$y^{\text{'}}=C_0\left(-\frac{3}{2}x+\frac{3}{8}{x}^2-\frac{1}{4}{x}^3-\frac{9}{64}{x}^4+\dots \right)+C_1\left(1-\frac{3}{4}{x}^2+\frac{1}{4}{x}^3+\dots \right)----\left(8\right)$

The initial conditions$y\left(0\right)=0$ and${\mathrm{y}}^{\text{'}}\left(0\right)=1$ give us${\mathrm{C}}_0=0$ and${\mathrm{C}}_1=1.$ Therefore, the solution of the given IVP is

$\mathrm{y}=\mathrm{x}-\frac{1}{4}{\mathrm{x}}^3+\frac{1}{16}{\mathrm{x}}^4+\dots$