Solving ordinary differential equation $x^{\prime\prime}+x=\sin t+e^{2t} $

Question

The question is: $$x^{\prime\prime}+x=\sin t+e^{2t} $$ I have tried to solve it like this: $$r^2+1=0 \implies r=\pm i $$ The answer to the homogeneous equation is: $$x_h(t)=C_1\cos t + C_2\sin t $$ but I don't know how should I find the particular solution I have tried with $$ x_p=A\cos t + B\sin t + Ce^{2t} $$ but I didn't get the right answer, any suggestion would be great.

Thanks.

Answer 1

$$x^{\prime\prime}+x=\sin t+e^{2t}$$ For the particular solution, you should simply try: $$x_p=At\cos t+Be^{2t}$$

Answer 2

You can find the particular solution as follows $$x_p=\frac{\sin t+e^{2t}}{D^2+1}\quad (\text{where}, \ D=\frac{d}{dt})$$ $$=\frac{\sin t}{D^2+1}+\frac{e^{2t}}{D^2+1}$$ $$=t\frac{\sin t}{\frac{d}{dD}(D^2+1)}+\frac{e^{2t}}{2^2+1}\quad (\because \ -1^2+1=0)$$ $$=t\frac{\sin t}{2D}+\frac{e^{2t}}{5}$$ $$=\frac{t}{2}D\left(\frac{\sin t}{D^2}\right)+\frac{e^{2t}}{5}$$ $$=\frac{t}{2}D\left(\frac{\sin t}{-1^2}\right)+\frac{e^{2t}}{5}$$ $$=\color{blue}{-\frac{t}{2}\cos t+\frac{e^{2t}}{5}}$$

Note: Standard formula used to find particular solutions

$$\boxed{x_p=\frac{\sin at}{D^2+b}=\frac{\sin at}{-a^2+b}(\text{if}\ -a^2+b\ne0)}$$ $$\boxed{x_p=\frac{\sin at}{D^2+b}=t\frac{\sin at}{\frac{d}{dD}(D^2+b)}\ \ (\text{if}\ -a^2+b=0)}$$ $$\boxed{x_p=\frac{e^{at}}{D^2+b}=\frac{e^{at}}{a^2+b}}$$