Solve the differential equation using Taylor-series expansion:

$$ \frac{dy}{dx} = x + y + xy \\ y (0) = 1 $$ to get value of $y$ at $x = 0.1$ and $x = 0.5$. Use terms through $x^5$.


The approach that I have seen taken when asked to determine the solution using Taylor's Theorem is as follows.

We have, from Taylor's Theorem, $$y(x)=y(0)+y'(0)x+\frac{y''(0)}{2}x^2+\frac{y^{(3)}(0)}{6}x^3+\ldots$$ which we need to solve for the respective coefficients.

We are given $y(0)=1$. When $x=0$, the ODE must be satisfied. Then we must have $$ y'(0) = 0+1+0\cdot1=1.$$ Differentiating the ODE we get $$\frac{d^2y}{dx^2}=1+\frac{dy}{dx}+y+x\frac{dy}{dx}.$$ Using this, we get that, at $x=0$, $$y''(0)=1+1+1+0\cdot1=3.$$

Then, we have $$y(x)=1+x+\frac{3}{2}x^2+\frac{y^{(3)}(0)}{6}x^3+\ldots$$ Continuing in this fashion, you can get the value of $y^{(3)}(0)$ and higher derivatives at $x=0$, thus giving a solution to the original ODE.

Once you have the required terms, you can evaluate the function at $x=0.1$ and $x=0.5$.

  • $\begingroup$ Do you know any reference for this method and what is it called in the literature? thanks $\endgroup$ Apr 11 '15 at 18:43
  • $\begingroup$ @Algohi I don't think it has a name. Googling "solve differential equation with Taylor series" brings up a few results you might find helpful. $\endgroup$
    – Daryl
    Apr 11 '15 at 23:34
  • $\begingroup$ I did google also and got some results. I just thought if there is any publication about the method. thanks anyway $\endgroup$ Apr 12 '15 at 3:19

Suppose that $y$ has the Taylor series expansion about $x=0$ given by $$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots.$$ Because $y(0)=1$, we have $a_0=1$. Differentiate. We get $$\frac{dy}{dx}=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+\cdots.\tag{$1$}$$ Also, $$\begin{align}x+y+xy&=x+(1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots)\\&+(x+a_1x^2+a_2x^3+a_3x^4+a_4x^5+\cdots).\end{align}$$ In the expression above, gather like powers of $x$ together. We get $$x+y+xy=1+(2+a_1)x+(a_1+a_2)x^2+(a_2+a_3)x^3+(a_4+a_5)x^4+\cdots.\tag{$2$}$$ The expansions $(1)$ and $(2)$ must be identical. It follows that they have the same constant term, that is, that $a_1=1$.

The coefficients of $x$ in $(1)$ and $(2)$ must match. It follows that $2a_2=2+a_1=3$, and therefore $a_2=\frac{3}{2}$.

The coefficients of $x^2$ must match. It follows that $3a_3=a_1+a_2=\frac{5}{2}$, and therefore $a_3=\frac{5}{6}$.

Continue, finding $a_4$ and $a_5$. You have not been asked to find coefficients beyond $a_5$.

For the numerical calculations, just substitute the given values of $x$ in the expression $1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$, using the values of the $a_i$ that we have found.

  • 2
    $\begingroup$ This looks like solving using the standard series approach and isn't really utilising Taylor's Theorem to obtain the solution. This is the approach I would take to solve the problem as well since it is more general, but I don't think it is what is being asked. $\endgroup$
    – Daryl
    Sep 22 '12 at 6:25
  • $\begingroup$ @Daryl: Hard to know, the OP can decide which one looks more like the course notes. The question said Taylor series, not Taylor's theorem. $\endgroup$ Sep 22 '12 at 6:28
  • $\begingroup$ Agree. Taylor series just 'special' power series, in one way to describe it. $\endgroup$
    – Daryl
    Sep 22 '12 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.