Q13 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem, $\mathbf{x}\mathbf{\text{'}}\mathbf{+}\left(\sin t\right)\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The solution is, $\mathrm{x}\left(\mathrm{t}\right)=1-\frac{{\mathrm{t}}^2}{2}+\frac{{\mathrm{t}}^4}{6}-...$.

Step 1: Solve the given differential equation.

The given differential equation is,

 $\mathrm{x}\text{'}+(\sin \mathrm{t})\mathrm{x}=0$

From the above equation, p(t) = sin t, which is an analytic function thus, the equation does not have any singularities:

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

The power series expansion of sin t is,

$\mathrm{sint}=\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+...$

Step 2: Take the derivative of the general solution.

Taking the derivative of the general solution and substituting it in the equation along with the expression of the sin t, we get the expression’

$\begin{array}{l}\mathrm{x}\text{'}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}.{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}\\ \sum _{\mathrm{n}=1}^{\infty }{\mathrm{na}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}-1}+\left(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+...\right)\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}=0\end{array}$

Expanding the series represented by individual terms of the expression we get,

$({\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{t}+3{\mathrm{a}}_3{\mathrm{t}}^2+4{\mathrm{a}}_4{\mathrm{t}}^3+...)+(\mathrm{t}-\frac{{\mathrm{t}}^3}{3!}+\frac{{\mathrm{t}}^5}{5!}-\frac{{\mathrm{t}}^7}{7!}+...)({\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+...)=0$

Step 3: Multiplying individual.

Multiplying individual term of the series and taking the common coefficients

$\begin{array}{l}({\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{t}+3{\mathrm{a}}_3{\mathrm{t}}^2+4{\mathrm{a}}_4{\mathrm{t}}^3+...)+({\mathrm{a}}_0\mathrm{t}+{\mathrm{a}}_1{\mathrm{t}}^2+{\mathrm{a}}_2{\mathrm{t}}^3+{\mathrm{a}}_3{\mathrm{t}}^4+...)\\ +\left(-{\mathrm{a}}_0\frac{{\mathrm{t}}^3}{3!}--{\mathrm{a}}_1\frac{{\mathrm{t}}^4}{3!}-{\mathrm{a}}_2\frac{{\mathrm{t}}^5}{3!}-{\mathrm{a}}_3\frac{{\mathrm{t}}^6}{3!}+...\right)+\left(-{\mathrm{a}}_0\frac{{\mathrm{t}}^5}{5!}--{\mathrm{a}}_1\frac{{\mathrm{t}}^6}{5!}-{\mathrm{a}}_2\frac{{\mathrm{t}}^7}{5!}-{\mathrm{a}}_3\frac{{\mathrm{t}}^9}{5!}+...\right)+...=0\\ a_1+(2a_2+a_0)t+(3a_3+a_1)t^2+\left(4a_4+a_2-\frac{a_0}{3!}\right)t^3+...=0\end{array}$

Equating coefficient equal to 0:

$\begin{array}{l}{\mathrm{a}}_1=0\\ 2{\mathrm{a}}_2+{\mathrm{a}}_0=0\Rightarrow {\mathrm{a}}_2=-\frac{{\mathrm{a}}_0}{2}\\ 3{\mathrm{a}}_3+{\mathrm{a}}_1=0\Rightarrow {\mathrm{a}}_3=0\\ 4{\mathrm{a}}_4+{\mathrm{a}}_2-\frac{{\mathrm{a}}_0}{6}=0\Rightarrow {\mathrm{a}}_4=\frac{{\mathrm{a}}_0}{6}\end{array}$

The general solution is given by,

 $\mathrm{x}\left(\mathrm{t}\right)=\sum _{\mathrm{t}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2+{\mathrm{a}}_3{\mathrm{t}}^3+{\mathrm{a}}_4{\mathrm{t}}^4+...$

Substituting,

$\mathrm{x}\left(\mathrm{t}\right)=1-\frac{{\mathrm{t}}^2}{2}+\frac{{\mathrm{t}}^4}{6}-...$

Hence the final solution is,

$\mathrm{x}\left(\mathrm{t}\right)=1-\frac{{\mathrm{t}}^2}{2}+\frac{{\mathrm{t}}^4}{6}-...$