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Linear Homogeneous ODEs : Complex Roots

    


Consider the linear homogeneous second order ODE \[ay''+by'+cy=0,\; a\neq 0. \;\;\;\;(35)\] In the last sections we discussed the cases when the roots \(r_1\) and \(r_2\) of the characteristic equation \(ar^2+br+c=0\) are real. Now assume \(b^2-4ac<0\). Then \(r_1\) and \(r_2\) are complex conjugate and they can be written as \(r_1=r+i\theta\) and \(r_2=r-i\theta\). Then we get two solutions \(y_1=e^{(r+i\theta)x}\) and \(y_2=e^{(r-i\theta)x}\). How to find two real fundamental solutions from \(y_1\) and \(y_2\)?

Answer: \(y=e^{rx}\cos(\theta x)\) and \(y=e^{rx}\sin(\theta x)\) are fundamental solutions of (35) and the general solution of (35) is \[y=c_1e^{rx}\cos(\theta x)+c_2e^{rx}\sin(\theta x).\]

We use Euler's formul: \(e^{i\theta}=\cos\theta+i\sin\theta\) for real \(\theta\).
Explanation: Recall from calculus that \[e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\] Applying this to \(e^{i\theta}\), we get \[\begin{align*} e^{i\theta}=\sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} &=1+\frac{i\theta}{1!}+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\cdots\\ &=1+\frac{i\theta}{1!}-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\cdots\\ &=\left(1-\frac{\theta^2}{2!}+\cdots\right) +i\left(\frac{\theta}{1!}-\frac{\theta^3}{3!}+\cdots\right)\\ &=\cos\theta+i\sin\theta. \end{align*}\] Using Eule'r formula we get \(y_1=e^{(r+i\theta)x}=e^{rx}e^{i\theta x}=e^{rx}(\cos(\theta x)+i\sin(\theta x))\) and similarly \(y_2=e^{(r-i\theta)x}=e^{rx}(\cos(\theta x)-i\sin(\theta x))\). Let's take their linear combination to get some real solutions: \[\begin{align*} y=d_1y_1+d_2y_2 &=d_1e^{rx}(\cos(\theta x)+i\sin(\theta x))+d_2e^{rx}(\cos(\theta x)-i\sin(\theta x))\\ &=(d_1+d_2)e^{rx}\cos(\theta x)+i(d_1-d_2)e^{rx}\sin(\theta x)\\ &=c_1e^{rx}\cos(\theta x)+c_2e^{rx}\sin(\theta x),\;\;\;\; \text{where } c_1=d_1+d_2,\,c_2=i(d_1-d_2) \end{align*}\] We can verify that \(y_1=e^{rx}\cos(\theta x)\) and \(y_2=e^{rx}\sin(\theta x)\) are solutions of (35). Let's see if they are linearly independent: \[\begin{align*} W(e^{rx}\cos(\theta x),\,e^{rx}\sin(\theta x)) &=\,\begin{array}{|cc|}e^{rx}\cos(\theta x)& e^{rx}\sin(\theta x)\\(e^{rx}\cos(\theta x))'&(e^{rx}\sin(\theta x))'\end{array}\\ &=\,\begin{array}{|cc|}e^{rx}\cos(\theta x)& e^{rx}\sin(\theta x)\\e^{rx}(r\cos(\theta x)-\theta\sin(\theta x))& e^{rx}(r\sin(\theta x)+\theta\cos(\theta x))\end{array}\\ &=e^{2rx}\theta(\cos^2(\theta x)+\sin^2(\theta x)) \\ &=e^{2rx}\theta \neq 0. \end{align*}\] Thus \(y_1=e^{rx}\cos(\theta x)\) and \(y_2=e^{rx}\sin(\theta x)\) are fundamental solutions of (35) and the general solution of (35) is \[y=c_1e^{rx}\cos(\theta x)+c_2e^{rx}\sin(\theta x).\]

Example. Find the general solution of the following ODE. \[\frac{d^2y}{dx^2}-6\frac{dy}{dx}+13y=0\] Solution. The characteristic equation is \[\begin{align*} r^2-6r+13&=0\\ r&=\frac{6\pm \sqrt{(-6)^2-4\cdot1\cdot 13}}{2\cdot 1}\\ r&=3+2i,3-2i \end{align*}\] So the general solution is \[y=c_1e^{3x}\cos(2x)+c_2e^{3x}\sin(2x).\]
Example. Solve the following IVP. \[y''+2y'+10y=0,\;\; y(0)=4,\, y'(0)=-7\] Solution. The characteristic equation is \[\begin{align*} r^2+2r+10&=0\\ (r+1)^2&=-9\\ r&=-1+3i,-1-3i. \end{align*}\] So the general solution is \[y=c_1e^{-x}\cos(3x)+c_2e^{-x}\sin(3x).\] Using the initial condition \(y(0)=4\), we get \[\begin{equation} c_1=4. \;\;\;\;(36) \end{equation}\] Note that \(y'=c_1(-\cos(3x)-3\sin(3x))e^{-x}+c_2(-\sin(3x)+3\cos(3x))e^{-x}.\) Using the initial condition \(y'(0)=-7\), we get \[\begin{equation} -c_1+3c_2=-7. \;\;\;\;(37) \end{equation}\] Solving (36) and (37), we get \(c_1=4\) and \(c_2=-1\). Thus the solution of the IVP is \[y=4e^{-x}\cos(3x)-e^{-x}\sin(3x).\]


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