Analytical methods: 2nd order

After dealing with first-order equations, we now look at the simplest type of second-order differential equation, with linear coefficients of the form ad2ydx2+bdydx+cy=0.

To solve this we look at the solutions to the auxiliary equation, given by ak2+bk+c=0.

Based on the solutions of the auxiliary equation, the solution takes the following forms:

  • If the solutions are k=α,β where α,β∈ℜ, with α≠β then the solutions to the original equation are of the form y=Aeαx+Beβx, where A,B∈ℜ are constants of integration to be determined by initial or boundary conditions.

  • If the solutions are both k=α where α∈ℜ then the solutions to the original equation are of the form y=(A+Bx)eαx, where A,B∈ℜ are constants of integration to be determined by initial or boundary conditions.

  • If the solutions are k=±βi where β∈ℜ then the solutions to the original equation are of the form y=Asin⁡(βx)+Bcos⁡(βx), where A,B∈ℜ are constants of integration to be determined by initial or boundary conditions.

  • If the solutions are k=α±βi where α,β∈ℜ then the solutions to the original equation are of the form y=eαx[Asin⁡(βx)+Bcos⁡(βx)], where A,B∈ℜ are constants of integration to be determined by initial or boundary conditions.

Walkthrough

We wish to solve d2ydx2+2dydx+y=0 subject to y(0)=0andy′(1)=1. The solutions of the auxiliary equation are both k=−1, therefore the general solution is y=(A+Bx)e−x, Applying the boundary conditions y(0)=0,y′(1)=1 gives B=e1,A=0, and the solution is therefore y(x)=xe1−x.