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 \[y=c_1e^{rx}\cos(\theta x)+c_2e^{rx}\sin(\theta x).\]