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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 $$$c_1=1. \;\;\;\;(33)$$$ Note that $$y'=-2c_1e^{-2x}+c_2(-2x+1)e^{-2x}.$$ Using the initial condition $$y'(0)=18$$, we get $$$-2c_1+c_2=18. \;\;\;\;(34)$$$ 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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