# Using Taylor series to solve differential equations

Up a level : Differential Equations
Previous page : First order linear differential equations - a couple of examples
Next page : Simple harmonic motion again – using Taylor series I bit further away along the pages we solve the equation $y' - xy = {x^2}$

You can have a look at it here. The problem we quickly encountered was that the solution is a non-elementary function. Let us tackle this in another way. Previously, in the section on power series we were able to solve the equation y´=y by assuming the solution could be expressed as a power series. Let us attempt something similar here.

Say we want a solution that goes through the point (1, 2). So we start with assuming we have a solution that can be expressed with a Taylor series, i.e. that $\begin{gathered} y = f(x) = \sum\limits_{k = 0}^\infty {\frac{{{f^{(k)}}(a)}}{{k!}}{{(x - a)}^k}} \hfill \\ \quad = f(a) + f'(a)(x - a) + \frac{{f''(a)}}{{2!}}{(x - a)^2} + \frac{{f'''(a)}}{{3!}}{(x - a)^3} + ... \hfill \\ \end{gathered}$

In this case we have that $y(1) = f(a) = f(1) = 2$

and that $y' = {x^2} + xy$

so $y'(1) = {1^2} + 1 \cdot 2 = 3 = f'(a)$

Differentiating our differential equation once gives us $y'' = 2x + y + xy'$

and thus $y''(1) = 2 \cdot 1 + 2 + 1 \cdot 3 = 7$

Differentiate again $y''' = 2 + y' + y' + y'' = 2 + 2y' + y''$

so $y'''(1) = 2 + 2 \cdot 3 + 7 = 15$

Differentiate again $\begin{gathered} {y^{(4)}} = 3y'' + y''', \hfill \\ {y^{(4)}}(1) = 3 \cdot 7 + 15 = 36 \hfill \\ \end{gathered}$

So, an approximate solution should be $y = 2 + 3(x - 1) + \frac{7}{2}{(x - 1)^2} + \frac{{15}}{6}{(x - 1)^3} + \frac{{36}}{{24}}{(x - 1)^4}$

In the graph below we can see the slope field, the particular solution in green (found in the above mentioned page) and the approximate solution in violet. If we try that for x=4 we get y=12.5, that is rather far from the about 13.96 we should get according to the exact solution. If we add some more terms we get $\begin{gathered} {y^{(5)}} = 4y'' + xy''', \hfill \\ {y^{(5)}}(1) = 3 \cdot 15 + 1 \cdot 36 = 96 \hfill \\ \end{gathered}$

From this point on we have that $\begin{gathered} {y^{(n)}} = (n - 1){y^{(n - 1)}} + x{y^n}, \hfill \\ {y^{(n)}}(1) = (n - 1){y^{(n - 1)}}(1) + {y^n}(1) \hfill \\ \end{gathered}$

This gives us the sequence 2, 3, 7, 15, 36, 96, 267, 852 and so on. We can use these values to get a better approximation.  Using our exact method we get 13.96225, using 15 terms in our series expansion we get 13.69221, so quite close. Up a level : Differential Equations
Previous page : First order linear differential equations - a couple of examples
Next page : Simple harmonic motion again – using Taylor series Last modified: Apr 1, 2019 @ 16:31