Linear equations of order n with constant coefficients

We have a non homogeneous linear ODE with constant coefficients. First of all, we solve the homogeneous part $$y''+4y=0$$ We have the characteristical polynomial: $p(\lambda)=\lambda^2+4$ that has roots:

• $\lambda=\pm2i$ complex root and its conjugate. The solution is then $y_1(x)=\cos(2x)$ and $y_1(x)=\sin(2x)$.

We look for a polynomial $Q(D)$ that cancels $f(x)$. To do this, we proceed the inverse way to the way we did it when computing the homogeneous solutions:

• $\cos(x)$ comes from a complex root $\lambda=i$

• $\sin(x)$ comes from a complex root $\lambda=-i$

• $1$ comes from a simple real root $\lambda=0$

Therefore the polynomial is $Q(D)=(D+Id\cdot i)(D-Id\cdot i)(D-0)=(D^2+Id)D$

We consider the new homogeneous problem $Q(D)P(D)y(x)=0$ $$Q(D)P(D)y(x)=(D^2+Id)D(D^2+4D)y(x)=$$ $$=(D+Id\cdot i)(D-Id\cdot i)D(D^2-4Id)y(x)=0$$ We see that the solutions to this problem are: $$y^*(x)=C_1\cos(2x)+C_2\sin(2x)+C_3+C_4\sin(x)+C_5\cos(x)$$ Now, we take the functions that are a solution of $y^*$, but were not of $y_h$, and look for a particular solution that is a linear combination of these solutions: $$y_p(x)=A+B\cdot \cos(x)+C\cdot\sin(x)$$ We designate this as a solution: $$\left. \begin {array} {l} y_p''+4y_p=4\cos(x)+3\sin(x)-8 \\ y_p''+4y_p=-B\cos(x)-C\sin(x)+4A+4B\cos(x)+4C\sin(x)= \\ = 4A+3B\cos(x)+3C\sin(x)\end{array}\right\}$$ $$\Rightarrow \left\{ \begin {array} {c} 4A=-8 \\ 3B=4 \\ 3C=3 \end{array}\right. \Rightarrow \left\{ \begin {array} {c} A=-2 \\ B=\dfrac{4}{3} \\ C=1 \end{array}\right.$$ Finally, we see that the general solution is: $$y(x)=y_h(x)+y_p(x)=C_1\cos(2x)+C_2\sin(2x)+\dfrac{4}{3}\cos(x)+\sin(x)-2$$