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Linear Homogeneous ODEs with Constant Coefficients


In this section, we learn how to solve a linear homogeneous second order ODE with constant coefficients. \[a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0, \text{ where } a\neq 0 \;\;\;\;(27)\] Suppose \(y=e^{rx}\) is a solution of (27) for some values of \(r\). Then \(y'=re^{rx}\) and \(y''=r^2e^{rx}\). Plugging these into (27), we get \[\begin{align*} ar^2e^{rx}+bre^{rx}+ce^{rx}&=0\\ e^{rx}(ar^2+br+c)&=0 \end{align*}\] Since \(e^{rx}\neq 0\), we get a quadratic equation in \(r\), called the characteristic equation of (27): \[ar^2+br+c=0 \;\;\;\;(28)\] There are three possibilities for the roots \(r_1\) and \(r_2\) of (28):

  1. \(r_1\) and \(r_2\) are real and distinct

  2. \(r_1\) and \(r_2\) are real and the same (see Section 3.3)

  3. \(r_1\) and \(r_2\) are complex conjugates (see Section 3.4)

For this section assume that \(r_1\) and \(r_2\) are real and distinct. Then we have two solutions \(y=e^{r_1x}\) and \(y=e^{r_2x}\) for (27). It can be verified that any of their linear combinations would also be a solution of (27) (principle of superposition). Thus we get the general solution: \[y=c_1e^{r_1x}+c_2e^{r_2x}\] Now \(c_1\) and \(c_2\) can be found using two initial conditions as discussed before.

Steps to solve:

  1. Find the characteristic equation \(ar^2+br+c=0\).

  2. Find the (distinct) roots \(r_1\) and \(r_2\).

  3. The general solution is \(y=c_1e^{r_1x}+c_2e^{r_2x}\).

  4. Find \(c_1\) and \(c_2\) when two initial conditions are given.

Example. Find the general solution of the following ODE. \[\frac{d^2y}{dx^2}-2\frac{dy}{dx}-3y=0\] Solution. The characteristic equation is \[\begin{align*} r^2-2r-3&=0\\ (r+1)(r-3)&=0\\ r&=-1,3 \end{align*}\] So the general solution is \[y=c_1e^{-x}+c_2e^{3x}.\]
Example. Solve the following IVP. \[2y''+5y'-3y=0,\;\; y(0)=11/7,\, y'(0)=-26/7\] Solution. The characteristic equation is \[\begin{align*} 2r^2+5r-3&=0\\ (2r-1)(r+3)&=0\\ r&=-3,1/2 \end{align*}\] So the general solution is \[y=c_1e^{-3x}+c_2e^{x/2}.\] Using the initial condition \(y(0)=11/7\), we get \[c_1+c_2=11/7. \;\;\;\;(29)\] Note that \(y'=-3c_1e^{-3x}+\frac{c_2}{2}e^{x/2}.\) Using the initial condition \(y'(0)=-26/7\), we get \[-3c_1+c_2/2=-26/7. \;\;\;\;(30)\] Solving (29) and (30), we get \(c_1=9/7\) and \(c_2=2/7\). Thus the solution of the IVP is \[y=\frac{9}{7} e^{-3x}+\frac{2}{7} e^{x/2}.\]

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