Maclaurin Series from Differential Equations (DP IB Analysis & Approaches (AA): HL): Revision Note

Maclaurin series for differential equations

Can I apply Maclaurin Series to solving differential equations?

• If you have a differential equation of the form $\frac{\mathrm{d}y}{\mathrm{d}x}=g(x,y)$ along with the value of $y\left(0\right)$ it is possible to build up the Maclaurin series of the solution $y=f\left(x\right)$ term by term

  • This does not necessarily tell you the explicit function of $x$ that corresponds to the Maclaurin series you are finding

  • But the Maclaurin series you find is the exact Maclaurin series for the solution to the differential equation

• The Maclaurin series can be used to approximate the value of the solution y = f(x) for different values of $x$

  • You can increase the accuracy of this approximation by calculating additional terms of the Maclaurin series for higher powers of $x$

How can I find the Maclaurin Series for the solution to a differential equation?

• STEP 1
  Use implicit differentiation to find expressions for $y\text{'}\text{'},y\text{'}\text{'}\text{'}$ etc., in terms of $x,y$ and lower-order derivatives of $y$ 

  • The number of derivatives you need to find depends on how many terms of the Maclaurin series you want to find

  • For example, if you want the Maclaurin series up to the $x^4$ term, then you will need to find derivatives up to $y^{\left(4\right)}$ (the fourth derivative of $y$)

• STEP 2
  Using the given initial value for $y\left(0\right)$, find the values of $y\text{'}\left(0\right),y\text{'}\text{'}\left(0\right),y\text{'}\text{'}\text{'}\left(0\right),$ etc., one by one 

  • Each value you find will then allow you to find the value for the next higher derivative

• STEP 3
  Put the values found in STEP 2 into the general Maclaurin series formula

  
  $f\left(x\right)=f\left(0\right)+x{f}^{\text{'}}\left(0\right)+\frac{x^2}{2!}{f}^{\text{'}\text{'}}\left(0\right)+...$
  

  • This formula is in your exam formula booklet

  • $y=f\left(x\right)$ is the solution to the differential equation, so $y\left(0\right)$ corresponds to $f\left(0\right)$ in the formula, $y\text{'}\left(0\right)$ corresponds to $f^{\text{'}}\left(0\right)$, and so on

• STEP 4
  Simplify the coefficients for each of the powers of $x$ in the resultant Maclaurin series