# Solving Differential Equations with Taylor Series

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Many differential equations don’t have solutions that can be expressed in terms of finite combinations of familiar functions. However, we can often solve for the Taylor series of the solution.

## Demonstration

For example, to solve the differential equation $y’^\prime+xy’+xy=e^x$, we can substitute the Taylor series $y=\sum_{n=0}^\infty a_n x^n$ and $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ and solve for the coefficients $a_n$.

Differentiating, we have $y’=\sum_{n=1}^\infty na_n x^{n-1}$ and $y’^\prime = \sum_{n=2}^\infty n(n-1) a_n x^{n-2}$. Substituting the derivatives in the differential equation, re-indexing so that all exponents are $n$, expressing all sums with the same starting index, and combining terms under a single sum, we condense the expression into a single polynomial.

\begin{align*} e^x &= y'' + xy' + xy \\ \sum_{n=0}^\infty \frac{x^n}{n!} &= \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + x\sum_{n=1}^\infty na_n x^{n-1} + x \sum_{n=0}^\infty a_n x^n \\ \sum_{n=0}^\infty \frac{x^n}{n!} &= \sum_{n=2}^\infty n(n-1)a_n x^{n-2} + \sum_{n=1}^\infty na_n x^{n} + \sum_{n=0}^\infty a_nx^{n+1} \\ \sum_{n=0}^\infty \frac{x^n}{n!} &= \sum_{n=0}^\infty (n+2)(n+1)a_{n+2} x^{n} + \sum_{n=1}^\infty na_n x^{n} + \sum_{n=1}^\infty a_{n-1}x^{n} \\ 1+\sum_{n=1}^\infty \frac{x^n}{n!} &= 2a_2 + \sum_{n=1}^\infty (n+2)(n+1)a_{n+2} x^{n} + \sum_{n=1}^\infty na_n x^{n} + \sum_{n=1}^\infty a_{n-1}x^{n} \\ 0 &= 2a_2-1 + \sum_{n=1}^\infty \left[ (n+2)(n+1)a_{n+2} + na_n + a_{n-1} - \frac{1}{n!} \right] x^{n} \end{align*}

For the expression to evaluate to $0$, we must have $2a_2-1=0$ and $(n+2)(n+1)a_{n+2} + na_n + a_{n-1} - \frac{1}{n!} = 0$ for $n > 1$. So, we can choose $a_0$ and $a_1$ to be our constants $a_0 = C_1$ and $a_1 = C_2$, set $a_2 = \frac{1}{2}$, and express all other coefficients $a_n$ for $n>3$ in terms of the constants $a_0 = C_1$ and $a_1 = C_2$ through a recurrence:

\begin{align*} 0 &= (n+2)(n+1)a_{n+2}+na_n+a_{n-1} - \frac{1}{n!} \hspace{1.3cm} ( n \geq 1) \\ a_{n+2} &= \frac{1}{(n+2)(n+1)} \left( \frac{1}{n!}-na_n-a_{n-1} \right) \hspace{1.8cm} (n \geq 1) \\ a_{n} &= \frac{1}{n(n-1)} \left( \frac{1}{(n-2)!} - (n-2)a_{n-2}-a_{n-3} \right) \hspace{.48cm} ( n \geq 3) \end{align*}

Thus, our solution is given by $y=\sum_{n=0}^\infty a_n x^n$ where $a_0=C_1$, $a_1=C_2$, $a_2=\frac{1}{2}$, and

\begin{align*} a_n = \frac{1}{n(n-1)} \left( \frac{1}{(n-2)!} - (n-2)a_{n-2}-a_{n-3} \right) \hspace{.5cm} (n \geq 3). \end{align*}

As another example, we will solve the differential equation $y’’^\prime=y’y$ using the same process. We write the solution as the Taylor series $y=\sum_{n=0}^\infty a_n x^n$, substitute its derivatives into the equation, and simplify.

\begin{align*} y''' &= y'y \\ \sum_{n=3}^\infty n(n-1)(n-2)a_n x^{n-3} &= \left( \sum_{n=1}^\infty na_nx^{n-1} \right) \left( \sum_{n=0}^\infty a_nx^n \right) \\ \sum_{n=3}^\infty n(n-1)(n-2)a_n x^{n-3} &= \left( \sum_{k=0}^\infty (k+1) a_{k+1} x^{k} \right) \left( \sum_{m=0}^\infty a_mx^m \right) \\ \sum_{n=3}^\infty n(n-1)(n-2)a_n x^{n-3} &= \sum_{k,m=0}^\infty (k+1) a_{k+1} a_m x^{k+m} \\ \sum_{n=0}^\infty (n+3)(n+2)(n+1)a_{n+3} x^{n} &= \sum_{n=0}^\infty \left[ \sum_{k=0}^n (k+1) a_{k+1} a_{n-k} \right] x^n \end{align*}

We can choose $a_0 = C_1$, $a_1 = C_2$, and $a_2 = C_3$ as our constants and express all other coefficients $a_n$ for $n \geq 3$ in terms of these constants through a recurrence:

\begin{align*} (n+3)(n+2)(n+1)a_{n+3} &= \sum_{k=0}^n (k+1)a_{k+1}a_{n-k} \hspace{1.4cm} (n \geq 0) \\ a_{n+3} &= \frac{ \sum_{k=0}^n (k+1)a_{k+1}a_{n-k} }{(n+3)(n+2)(n+1)} \hspace{.9cm} (n \geq 0) \\ a_{n} &= \frac{ \sum_{k=0}^{n-3} (k+1)a_{k+1}a_{n-3-k} }{n(n-1)(n-2)} \hspace{.5cm} (n \geq 3) \end{align*}

Thus, our solution is given by $y=\sum_{n=0}^\infty a_n x^n$ where $a_0=C_1$, $a_1=C_2$, $a_2 = C_3$, and

\begin{align*} a_n = \frac{\sum_{k=0}^{n-3} (k+1)a_{k+1}a_{n-3-k} }{n(n-1)(n-2)} \hspace{.5cm} (n \geq 3). \end{align*}

## Exercises

Use Taylor series to solve the following differential equations. (You can view the solution by clicking on the problem.)

\begin{align*} 1) \hspace{.5cm} y'''+x^2y=1 \end{align*}
Solution:
\begin{align*} & a_0 = C_1, \, a_1=C_2, \, a_2 = C_3, \, a_3 = \frac{1}{6} , \, a_4=0 \\ & a_n = -\frac{a_{n-5} }{n(n-1)(n-2)} \hspace{.5cm} (n \geq 5) \end{align*}

\begin{align*} 2) \hspace{.5cm} y''+xy=e^x \end{align*}
Solution:
\begin{align*} & a_0 = C_1, \, a_1=C_2, \, a_2 = \frac{1}{2} \\ & a_n = \frac{1}{n(n-1)} \left[ \frac{1}{(n-2)!} - a_{n-3} \right] \hspace{.5cm} (n \geq 3) \end{align*}

\begin{align*} 3) \hspace{.5cm} y'''=\frac{y'}{1-x} \end{align*}
Solution:
\begin{align*} & a_0 = C_1, \, a_1=C_2, \, a_2 = C_3 \\ & a_n = \frac{ \sum_{k=0}^{n-3} (k+1)a_{k+1} }{n(n-1)(n-2)} \hspace{.5cm} (n \geq 3) \end{align*}

\begin{align*} 1) \hspace{.5cm} y'''=xy' \ln(1+x) \end{align*}
Solution:
\begin{align*} & a_0 = C_1, \, a_1=C_2, \, a_2 = C_3, \, a_3=0, \, a_4=0 \\ & a_n = \frac{ \sum_{k=1}^{n-4} \frac{(-1)^{n-k}}{n-k-3} a_k }{n(n-1)(n-2)} \hspace{.5cm} (n \geq 5) \end{align*}

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