Series Solutions to Differential Equations (Edexcel A Level Further Maths: Further Pure 1): Revision Note

You will be given the Taylor series expansion below within the exam question itself. For differential equation solutions it helps to swap the $\mathrm{f}$ to a $y$.

$\mathrm{f}\left(x\right)=\mathrm{f}\left(a\right)+\left(x-a\right)\mathrm{f}\text{'}\left(a\right)+\frac{{\left(x-a\right)}^2}{2!}\mathrm{f}\text{'}\text{'}\left(a\right)+...+\frac{{\left(x-a\right)}^r}{r!}{\mathrm{f}}^{\left(r\right)}\left(a\right)+...$

Make sure you know the difference between powers of derivatives and higher-order derivatives.

• E.g. ${\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2$ is the first derivative of $y$ with respect to $x$, squared

• $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}$ is the second derivative of $y$ with respect to $x$

• They are not the same thing!

The Taylor series expansion of $\mathrm{f}\left(x\right)$ about $x=a$ is given by

$\mathrm{f}\left(x\right)=\mathrm{f}\left(a\right)+\left(x-a\right)\mathrm{f}\text{'}\left(a\right)+\frac{{\left(x-a\right)}^2}{2!}\mathrm{f}\text{'}\text{'}\left(a\right)+...+\frac{{\left(x-a\right)}^r}{r!}{\mathrm{f}}^{\left(r\right)}\left(a\right)+...$

A differential equation is given by

$y^3\frac{\mathrm{d}y}{\mathrm{d}x}=\ln y+x^2$

where $y=1$ at $x=2$.

Determine a series solution for $y$, in ascending powers of $\left(x-2\right)$, up to and including the term in ${\left(x-2\right)}^2$, giving each coefficient in simplest form.

Answer:

Write out the relevant terms of the Taylor series formula, changing $\mathrm{f}$ into $y$ and substituting in $a=2$

$y\left(x\right)=y\left(2\right)+\left(x-2\right)y\text{'}\left(2\right)+\frac{{\left(x-2\right)}^2}{2!}y\text{'}\text{'}\left(2\right)+...$

Find $y\left(2\right)$, the value of $y$ when $x=2$ (it is given in the question)

$y\left(2\right)=1$

Next find $y\text{'}\left(2\right)$

First substitute $x=2$ and $y=1$ into the original differential equation

$1^3\frac{\mathrm{d}y}{\mathrm{d}x}=\ln 1+2^2$

Then make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject (this gives the value of $y\text{'}\left(2\right)$)

$\begin{array}{rcl}\frac{\mathrm{d}y}{\mathrm{d}x}& =& 0+4\\ y\text{'}\left(2\right)& =& 4\end{array}$

To find $y\text{'}\text{'}\left(2\right)$, don't differentiate the version with numbers substituted in

Instead, go back to the original differential equation and differentiate both sides with respect to $x$

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\left(y^3\frac{\mathrm{d}y}{\mathrm{d}x}\right)& =& \frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y+x^2\right)\end{array}$

It is easier here to start on the right-hand side, which can be separated as follows

$\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)\end{array}$

To calculate $\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y\right)\end{array}$, use implicit differentiation

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y\right)& =& \frac{\mathrm{d}}{\mathrm{d}y}\left(\ln y\right)\times \frac{\mathrm{d}y}{\mathrm{d}x}\\ & =& \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}\end{array}$

To calculate $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)$, differentiate $x^2$ with respect to $x$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right)=2x$

Now look at the left-hand side, which is the derivative of a product of terms

$\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(y^3\frac{\mathrm{d}y}{\mathrm{d}x}\right)\end{array}$

Apply the product rule, showing clearly the derivatives with respect to $x$

$\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(y^3\right)\frac{\mathrm{d}y}{\mathrm{d}x}+y^3\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\end{array}$

Use implicit differentiation to simplify the part $\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(y^3\right)\end{array}$

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\left(y^3\right)& =& \frac{\mathrm{d}}{\mathrm{d}y}\left(y^3\right)\times \frac{\mathrm{d}y}{\mathrm{d}x}\\ & =& 3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\end{array}$

Also simplify the derivative of a derivative, $\begin{array}{rcl}& & \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\end{array}$

$\begin{array}{rcl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)& =& \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\end{array}$

Substituting these parts back into the left-hand side

$\begin{array}{rcl}& & 3y^2\frac{\mathrm{d}y}{\mathrm{d}x}\times \frac{\mathrm{d}y}{\mathrm{d}x}+y^3\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\end{array}$

Simplify the first term using that $\frac{\mathrm{d}y}{\mathrm{d}x}\times \frac{\mathrm{d}y}{\mathrm{d}x}={\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2$

$\begin{array}{rcl}& & 3y^2{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2+y^3\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\end{array}$

Put the left-hand side and right-hand side back together

$\begin{array}{rcl}3y^2{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2+y^3\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{1}{y}\frac{\mathrm{d}y}{\mathrm{d}x}+2x\end{array}$

To find $y\text{'}\text{'}\left(2\right)$, substitute $x=2$ and $y=1$ into the equation above and make $\begin{array}{rcl}& & \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}\end{array}$ the subject

• ${\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}_{x=2}=y\text{'}\left(2\right)=4$, as found above

$\begin{array}{rcl}3\times 1^2{\left(4\right)}^2+1^3\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& \frac{1}{1}\times 4+2\times 2\\ 48+\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& 4+4\\ \frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}& =& -40\\ y\text{'}\text{'}\left(2\right)& =& -40\end{array}$

Finally, substitute $y\left(2\right)=1$, $y\text{'}\left(2\right)=4$ and $y\text{'}\text{'}\left(2\right)=-40$ into the Taylor series

$\begin{array}{rcl}y\left(x\right)& =& y\left(2\right)+\left(x-2\right)y\text{'}\left(2\right)+\frac{{\left(x-2\right)}^2}{2!}y\text{'}\text{'}\left(2\right)+...\\ & =& 1+\left(x-2\right)\times 4+\frac{{\left(x-2\right)}^2}{2\times 1}\times \left(-40\right)+...\\ & =& 1+4\left(x-2\right)-20{\left(x-2\right)}^2+...\end{array}$

Write out the series solution for $y$, up to and including the term in ${\left(x-2\right)}^2$

$\begin{array}{rcl}y& =& 1+4\left(x-2\right)-20{\left(x-2\right)}^2\end{array}$

Don't forget to put $y=$ at the start of your series solution, as this is an approximation to the exact curve $y$ that satisfies the differential equation.