Series solutions of differential equationsEdexcel International A Level Further Maths: Revision notes

A differential equation with initial conditions at $x=0$ can be solved as a series $y=y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+\frac{x^3}{3!}y'''(0)+\dots$ (the Maclaurin series of $y$), without finding a closed-form solution. The method is:

1. Use the initial conditions for $y(0)$ (and $y'(0)$ if second order).
2. Use the equation itself at $x=0$ to find the next derivative.
3. Differentiate the equation with respect to $x$ (product rule, chain rule) to get an expression for the next derivative, and evaluate it at $x=0$.
4. Repeat until you have all derivatives needed, then substitute into the series. For a first-order equation $y'=f(x,y)$ only $y(0)$ is needed. For a second-order equation both $y(0)$ and $y'(0)$ are needed.

Example: $\frac{dy}{dx}=x^2+y$ with $y(0)=1$.

• $y'(0)=0+1=1$.
• Differentiate: $y''=2x+y'$, so $y''(0)=1$.
• Differentiate: $y'''=2+y''$, so $y'''(0)=3$.
• Differentiate: $y^{(4)}=y'''$, so $y^{(4)}(0)=3$. So $y=1+x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{8}+\dots$ At $x=0.2$ this gives $1.2242$. (The exact solution is $y=3\mathrm{e}^x-x^2-2x-2$, which gives $1.2242$ to 4 d.p.)

Example (specification): $\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0$ with $y(0)=1$, $y'(0)=0$. The equation gives $y''=-xy'-y$, so $y''(0)=-1$. Differentiating, $y'''=-y'-xy''-y'=-2y'-xy''$, so $y'''(0)=0$. Differentiating again, $y^{(4)}=-2y''-y''-xy'''=-3y''-xy'''$, so $y^{(4)}(0)=3$. Hence $y=1-\frac{x^2}{2}+\frac{3x^4}{24}=1-\frac{x^2}{2}+\frac{x^4}{8}+\dots$ Another example: $y''=xy$ with $y(0)=1$, $y'(0)=1$ gives $y'''=y+xy'$, $y^{(4)}=2y'+xy''$, so $y=1+x+\frac{x^3}{6}+\frac{x^4}{12}+\dots$

Terms such as $y^2$ or $yy'$ need the chain rule and product rule when differentiated: $\frac{d}{dx}(y^2)=2yy'$ and $\frac{d}{dx}(yy')=y'^2+yy''$. Example: $y''=xy'+y^2$ with $y(0)=1$, $y'(0)=1$.

• $y''(0)=0+1=1$.
• $y'''=y'+xy''+2yy'$, so $y'''(0)=1+0+2=3$.
• $y^{(4)}=2y''+xy'''+2y'^2+2yy''$, so $y^{(4)}(0)=2+0+2+2=6$. Hence $y=1+x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{4}+\dots$

Substitute a small value of $x$ to estimate $y$. The accuracy improves for smaller $|x|$ and with more terms: for $y=1+2x+x^2+\frac56x^3+\frac16x^4$, $y(0.2)\approx1.4469$. Check: differentiate the series and substitute back into the differential equation. All terms up to the order you have computed must cancel. For $y''=xy'+y^2$ with $y=1+x+\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^4}{4}$: $y''=1+3x+3x^2+\dots$, $xy'=x+x^2+\dots$ and $y^2=1+2x+2x^2+\dots$, so $y''-xy'-y^2=0$ up to $x^2$. If the conditions are given at $x=a$ rather than $x=0$, use the Taylor series in powers of $(x-a)$ in the same way.