24. Applications of Taylor Series

e. Differential Equations

Finally, power series can be used to find approximate solutions to initial value problems, even when the differential equation is neither separable nor linear.

Find the quadratic Taylor polynomial centered at \(x=0\), for the solution to the initial value problem \[ \dfrac{dy}{dx}=x^2+y^2 \qquad \text{with} \qquad y(0)=3. \]

The quadratic Taylor polynomial about \(x=0\) has the form: \[ y(x)\approx y(0)+y'(0)x+\dfrac{1}{2}y''(0)x^2 \] So all we need is to find \(y(0)\), \(y'(0)\) and \(y''(0)\). The first two come directly from the initial condition and the differential equation: \[\begin{aligned} y(0)&=3 \\ y'(0) &=\left.\dfrac{dy}{dx}\right|_{x=0} =\left[x^2+y^2\right]_{x=0} \\ &=0^2+y(0)^2=3^2=9 \end{aligned}\] The third comes from implicitly differentiating the differential equation: \[\begin{aligned} \dfrac{d^2y}{dx^2} &=\dfrac{d}{dx}(x^2+y^2)=2x+2y\dfrac{dy}{dx} \\ y''(0) &=\left[\dfrac{d^2y}{dx^2}\right]_{x=0} =\left[2x+2y\dfrac{dy}{dx}\right]_{x=0} \\ &=2y(0)y'(0)=2\cdot3\cdot9=54 \end{aligned}\] So the quadratic Taylor polynomial about \(x=0\) is: \[ y(x)\approx3+9x+27x^2 \] Don't forget the \(\dfrac{1}{2}\) in front of the quadratic term!

Obviously this process could be continued to produce the higher degree Taylor polynomials.