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

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,

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

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,

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

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,

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