Problems from Linear equations of order n with constant coefficients

Development:

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:

• $\lambda =\pm 2i$ 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: $$y^{\ast }(x)=C_1\cos (2x)+C_2\sin (2x)+C_3+C_4\sin (x)+C_5\cos (x)$$ Now, we take the functions that are a solution of $y^{\ast }$, 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: $$\begin{array}{l}{y}_p^{″}+4y_p=4\cos (x)+3\sin (x)-8\\ y_p^{″}+4y_p=-B\cos (x)-C\sin (x)+4A+4B\cos (x)+4C\sin (x)=\\ =4A+3B\cos (x)+3C\sin (x)\end{array}\}$$ $$\Rightarrow \{\begin{array}{c}4A=-8\\ 3B=4\\ 3C=3\end{array}\Rightarrow \{\begin{array}{c}A=-2\\ B={\displaystyle \frac{4}{3}}\\ C=1\end{array}$$ Finally, we see that the general solution is: $$y(x)=y_h(x)+y_p(x)=C_1\cos (2x)+C_2\sin (2x)+{\displaystyle \frac{4}{3}}\cos (x)+\sin (x)-2$$

Solution:

$y(x)=C_1\cos (2x)+C_2\sin (2x)+{\displaystyle \frac{4}{3}}\cos (x)+\sin (x)-2$

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