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I have the solve the following exercise:

Exercise:

Describe the qualitative nature of the solution of the initial value problem

$$ y'' + 4y = g(t)\\ y(0) = 0, y'(0) = 0 $$ where $$ g(t)= \begin{cases} 0, & 0 \leq t < 5\\\dfrac{1}{5}(t - 5), & 5\leq t \leq 10\\1, & t \geq 10\end{cases} $$ and then find the solution.

In the book the following solution is given:

Solution: In this example the forcing function has the graph shown in Figure 6.4.2 and is known as ramp loading. Is is relatively easy to identify the general form of the solution. For $t<5$, the solution is simply $y = 0$. On the other hand, for $t>10$, the solution has the form $$ y = c_1\cos(2t) + c_2\sin(2t) + \dfrac{1}{4} $$ The constant $1/4$ is a particular solution of the nonhomogeneous equation, while the other two terms are the general solution of the corresponding homogeneous solution.

My Question: I tried to solve the homogeneous equation by assuming that the solution is of the form $y = e^{rt}$. In that case the characteristic equation becomes $r^2 + 4r = 0$ so that we have $r = 0$ or $r = -4$ and $y_c = c_1 + c_2e^{-4t}$. Is this incorrect? Why should $y_c$ be equal to $c_1\cos(2t) + c_2\sin(2t)$?

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  • $\begingroup$ You even know that in the middle part, the solution is $y=b_1\cos(2t)+b_2\sin(2t)+\frac14(t/5-1)$, as the second derivative of linear functions is zero. $\endgroup$ – Lutz Lehmann Feb 14 '19 at 11:54
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The characteristic equation should not be $r^2 + 4r=0$, but rather $r^2 +4=0$. This is because the LHS of the ODE is $y'' +4y$, i.e. $y'' +\color{blue}{0y'}+4y$. So the characteristic equation is $r^2 +\color{blue}{0r}+4 =0$, i.e. $r^2+4=0$.

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