Question

In Problems 13-19,find at least the first four non-zero terms in a power series expansion of the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

The first four nonzero terms in the power series expansion of the given initial value problem$\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}+{\mathrm{e}}^{\mathrm{t}}\mathrm{y}=0$is$\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^3}{3}+\frac{{\mathrm{t}}^4}{12}-\frac{{\mathrm{t}}^5}{24}$.

Step-by-step solution

Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

$\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Find the relation.

Given,

$\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}+{\mathrm{e}}^{\mathrm{t}}\mathrm{y}=0;\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=-1$

Use the formula,

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

Taking derivative and substituting in the equation, we get the relation,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1} \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2} \\ \sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+\mathrm{t}·\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}+{\mathrm{e}}^{\mathrm{t}}·\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0 \end{gathered}$$

Hence we get the relation $\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}(\mathrm{n}-1)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}+\mathrm{t}·\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}+{\mathrm{e}}^{\mathrm{t}}·\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$.

Find the expression after expansion.

The series expansion for the function is

$$\begin{gathered} \left(2{\mathrm{a}}_2+6{\mathrm{a}}_3\mathrm{t}+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3+30{\mathrm{a}}_6{\mathrm{t}}^4+\cdots \right)+\mathrm{t}·\left({\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{t}+3{\mathrm{a}}_3{\mathrm{t}}^2+4{\mathrm{a}}_4{\mathrm{t}}^4+5{\mathrm{a}}_5{\mathrm{t}}^4+\cdots \right) \\ +\left(1+\mathrm{t}+\frac{{\mathrm{t}}^2}{2}+\frac{{\mathrm{t}}^3}{6}+\cdots \right)\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+\cdots \right)=0 \end{gathered}$$

By expanding the series we get,

role="math" localid="1664095204500" $$\begin{gathered} \left(2{\mathrm{a}}_2+6{\mathrm{a}}_3\mathrm{t}+12{\mathrm{a}}_4{\mathrm{t}}^2+20{\mathrm{a}}_5{\mathrm{t}}^3+30{\mathrm{a}}_6{\mathrm{t}}^4+\cdots \right)+\left({\mathrm{a}}_1\mathrm{t}+2{\mathrm{a}}_2{\mathrm{t}}^2+3{\mathrm{a}}_3{\mathrm{t}}^3+4{\mathrm{a}}_4{\mathrm{t}}^5+5{\mathrm{a}}_5{\mathrm{t}}^6+\cdots \right) \\ +\left({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+\cdots \right)+\left({\mathrm{a}}_0\mathrm{t}+{\mathrm{a}}_1{\mathrm{t}}^2+{\mathrm{a}}_2{\mathrm{t}}^3+{\mathrm{a}}_3{\mathrm{t}}^4+\cdots \right)+\left({\mathrm{a}}_0\frac{{\mathrm{t}}^2}{2}\right.\left.+{\mathrm{a}}_1\frac{{\mathrm{t}}^3}{2}+{\mathrm{a}}_2\frac{{\mathrm{t}}^4}{2}+{\mathrm{a}}_3\frac{{\mathrm{t}}^5}{2}+\cdots \right) \\ +\left({\mathrm{a}}_0\frac{{\mathrm{t}}^3}{6}+{\mathrm{a}}_1\frac{{\mathrm{t}}^4}{6}+{\mathrm{a}}_2\frac{{\mathrm{t}}^5}{6}+{\mathrm{a}}_3\frac{{\mathrm{t}}^6}{6}+\cdots \right)+\cdots =0 \end{gathered}$$

Simplify the expression.

$$\begin{gathered} \left(2{\mathrm{a}}_2+{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3+{\mathrm{a}}_1+{\mathrm{a}}_1+{\mathrm{a}}_0\right)\mathrm{t}+\left(12{\mathrm{a}}_4+\right.\left.{\mathrm{a}}_2+{\mathrm{a}}_2+{\mathrm{a}}_1+\frac{{\mathrm{a}}_0}{2}\right){\mathrm{t}}^2 \\ +\left(20{\mathrm{a}}_5+3{\mathrm{a}}_3+{\mathrm{a}}_3+{\mathrm{a}}_2+\frac{{\mathrm{a}}_1}{2}+\frac{{\mathrm{a}}_0}{6}+\cdots \right){\mathrm{t}}^3+\cdots =0 \end{gathered}$$

Hence, the expression after the expansion is:

$$\begin{gathered} \left(2{\mathrm{a}}_2+{\mathrm{a}}_0\right)+\left(6{\mathrm{a}}_3+{\mathrm{a}}_1+{\mathrm{a}}_1+{\mathrm{a}}_0\right)\mathrm{t}+\left(12{\mathrm{a}}_4+\right.\left.{\mathrm{a}}_2+{\mathrm{a}}_2+{\mathrm{a}}_1+\frac{{\mathrm{a}}_0}{2}\right){\mathrm{t}}^2 \\ +\left(20{\mathrm{a}}_5+3{\mathrm{a}}_3+{\mathrm{a}}_3+{\mathrm{a}}_2+\frac{{\mathrm{a}}_1}{2}+\frac{{\mathrm{a}}_0}{6}+\cdots \right){\mathrm{t}}^3+\cdots =0 \end{gathered}$$

Find the first four nonzero terms.

By equating the coefficients, we get,

$$\begin{gathered} 6{\mathrm{a}}_3+{\mathrm{a}}_1+{\mathrm{a}}_1+{\mathrm{a}}_0=0\to {\mathrm{a}}_3=-\frac{{\mathrm{a}}_1}{3}=\frac{1}{3} \\ 12{\mathrm{a}}_4+{\mathrm{a}}_2+{\mathrm{a}}_2+{\mathrm{a}}_1+\frac{{\mathrm{a}}_0}{2}=0\to {\mathrm{a}}_4=\frac{-{\mathrm{a}}_0}{12}=\frac{1}{12} \\ 20{\mathrm{a}}_5+3{\mathrm{a}}_3+{\mathrm{a}}_3+{\mathrm{a}}_2+\frac{{\mathrm{a}}_1}{2}+\frac{{\mathrm{a}}_0}{6}\to {\mathrm{a}}_5=\frac{-4{\mathrm{a}}_3-\frac{{\mathrm{a}}_1}{2}}{20}=-\frac{1}{24} \end{gathered}$$

The general solution was

$\mathrm{y}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+\cdots$

Apply the initial condition and substitute the coefficient.

$\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^3}{3}+\frac{{\mathrm{t}}^4}{12}-\frac{{\mathrm{t}}^5}{24}$

Hence, the first four nonzero terms are$\mathrm{y}\left(\mathrm{t}\right)=-\mathrm{t}+\frac{{\mathrm{t}}^3}{3}+\frac{{\mathrm{t}}^4}{12}-\frac{{\mathrm{t}}^5}{24}$.