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Linear Homogeneous ODEs : Repeated Real Roots


Consider the linear homogeneous second order ODE \[ay''+by'+cy=0,\; a\neq 0. \;\;\;\;(32)\] We know that if the roots \(r_1\) and \(r_2\) of the characteristic equation \(ar^2+br+c=0\) are real and distinct, then \(y=e^{r_1x}\) and \(y=e^{r_2x}\) are fundamental solutions. Now assume \(b^2-4ac=0\). Then \(r_1=r_2=-b/2a\) and we get only one fundamental solution \(y=e^{r_1x}\). How to find the second fundamental solution?

Let \(y=v(x)e^{r_1x}\) be a solution of (32). Plugging \(y=ve^{r_1x}\), \(y'=v'e^{r_1x}+r_1ve^{r_1x}\) and \(y''=v''e^{r_1x}+2r_1v'e^{r_1x}+r_1^2ve^{r_1x}\) into (32), we get \[\begin{align*} a(v''e^{r_1x}+2r_1v'e^{r_1x}+r_1^2ve^{r_1x})+b(v'e^{r_1x}+r_1ve^{r_1x})+cve^{r_1x}&=0&\\ [av''+(2ar_1+b)v'+(ar_1^2+br_1+c)v]e^{r_1x}&=0&\\ av''+(2ar_1+b)v'+(ar_1^2+br_1+c)v&=0&(\text{since } e^{r_1x}\neq 0)\\ av''+(2ar_1+b)v'&=0& (\text{since } ar_1^2+br_1+c=0)\\ av''&=0& (\text{since } r_1=-b/2a, 2ar_1+b=0)\\ v''&=0& (\text{since } a\neq 0) \end{align*}\] Then \(v'(x)=c_2\), a constant and \(v(x)=\int v'(x)\, dx=\int c_2\, dx=c_2x+c_1\). Thus \(y=(c_1+c_2x)e^{r_1x}=c_1e^{r_1x}+c_2xe^{r_1x}\). We can verify that \(y_1=e^{r_1x}\) and \(y_2=xe^{r_1x}\) are solutions of (32). This technique of getting another solution from a given solution is called reduction of order. Let's see if they are linearly independent: \[\begin{align*} W(e^{r_1x},\,xe^{r_1x}) =\,\begin{array}{|cc|}e^{r_1x}&xe^{r_1x}\\(e^{r_1x})'&(xe^{r_1x})'\end{array} =\,\begin{array}{|cc|}e^{r_1x}&xe^{r_1x}\\r_1e^{r_1x}&(r_1x+1)e^{r_1x}\end{array} &=e^{r_1x}\cdot (r_1x+1)e^{r_1x}-xe^{r_1x}\cdot r_1e^{r_1x}\\&=e^{2r_1x} \neq 0. \end{align*}\] Thus \(y_1=e^{r_1x}\) and \(y_2=xe^{r_1x}\) are fundamental solutions of (32) and the general solution of (32) is \[y=c_1e^{r_1x}+c_2xe^{r_1x}.\]

Example. Find the general solution of the following ODE. \[\frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0\] Solution. The characteristic equation is \[\begin{align*} r^2-6r+9&=0\\ (r-3)^2&=0\\ r&=3,3 \end{align*}\] So the general solution is \[y=c_1e^{3x}+c_2xe^{3x}.\]
Example. Solve the following IVP. \[y''+4y'+4y=0,\;\; y(0)=1,\, y'(0)=18\] Solution. The characteristic equation is \[\begin{align*} r^2+4r+4&=0\\ (r+2)^2&=0\\ r&=-2,-2. \end{align*}\] So the general solution is \[y=c_1e^{-2x}+c_2xe^{-2x}.\] Using the initial condition \(y(0)=1\), we get \[\begin{equation} c_1=1. \;\;\;\;(33) \end{equation}\] Note that \(y'=-2c_1e^{-2x}+c_2(-2x+1)e^{-2x}.\) Using the initial condition \(y'(0)=18\), we get \[\begin{equation} -2c_1+c_2=18. \;\;\;\;(34) \end{equation}\] Solving (33) and (34), we get \(c_1=1\) and \(c_2=20\). Thus the solution of the IVP is \[y=e^{-2x}+20xe^{-2x}.\]

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