How do I find out particular solution for my differential equation? How do I find out particular solution for my differential equation?
For example $\ddot{x}+4\dot{x}+4x=t+1+\sin t$.
Can someone explain me why is particular solution here $x_p(t)=At+B+C\sin t+D\cos t$?

Use Laplace transform:
$$x''(t)+4x'(t)+4x(t)=1+t+\sin(t)\Longleftrightarrow$$
$$\mathcal{L}_t\left[x''(t)+4x'(t)+4x(t)\right]_{(s)}=\mathcal{L}_t\left[1+t+\sin(t)\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_t\left[x''(t)\right]_{(s)}+\mathcal{L}_t\left[4x'(t)\right]_{(s)}+\mathcal{L}_t\left[4x(t)\right]_{(s)}=\mathcal{L}_t\left[1\right]_{(s)}+\mathcal{L}_t\left[t\right]_{(s)}+\mathcal{L}_t\left[\sin(t)\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_t\left[x''(t)\right]_{(s)}+4\cdot\mathcal{L}_t\left[x'(t)\right]_{(s)}+4\cdot\mathcal{L}_t\left[x(t)\right]_{(s)}=\mathcal{L}_t\left[1\right]_{(s)}+\mathcal{L}_t\left[t\right]_{(s)}+\mathcal{L}_t\left[\sin(t)\right]_{(s)}\Longleftrightarrow$$

Now, use:


*

*$$\mathcal{L}_t\left[1\right]_{(s)}=\frac{1}{s}$$

*$$\mathcal{L}_t\left[t\right]_{(s)}=\frac{1}{s^2}$$

*$$\mathcal{L}_t\left[\sin(t)\right]_{(s)}=\frac{1}{1+s^2}$$

*$$\mathcal{L}_t\left[x(t)\right]_{(s)}=\text{X}(s)$$

*$$\mathcal{L}_t\left[x'(t)\right]_{(s)}=s\text{X}(s)-x(0)$$

*$$\mathcal{L}_t\left[x''(t)\right]_{(s)}=s^2\text{X}(s)-sx(0)-x'(0)$$


$$\left(s^2\text{X}(s)-sx(0)-x'(0)\right)+\left(4s\text{X}(s)-4x(0)\right)+\left(4\text{X}(s)\right)=\left(\frac{1}{s}\right)+\left(\frac{1}{s^2}\right)+\left(\frac{1}{1+s^2}\right)\Longleftrightarrow$$
$$\text{X}(s)=\frac{\frac{1}{s}+\frac{1}{s^2}+\frac{1}{1+s^2}+4x(0)+sx(0)+x'(0)}{(s+2)^2}$$
Now, with inverse Laplace transform we find:
$$x(t)=\frac{25t-16\cos(t)+12\sin(t)+e^{-2t}\left(4(4+25x(0))+5t(40x(0)+20x'(0)-1)\right)}{100}$$
