DiffEq Home

## 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 $$$c_1=4. \;\;\;\;(36)$$$ 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 $$$-c_1+3c_2=-7. \;\;\;\;(37)$$$ Solving (36) and (37), we get $$c_1=4$$ and $$c_2=-1$$. Thus the solution of the IVP is $y=4e^{-x}\cos(3x)-e^{-x}\sin(3x).$

Last edited