Question

# Find all real solutions o f the differential equations. $\frac{d^{2} x}{d t^{2}}+2 x=\cos (t)$

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Consider the differential equation

$\frac{d^2x}{dt^2}+2x=\cos(t)$

The characteristic polynomial of the operator $T(x)=\frac{d^2x}{dt^2}+2x$ is

$p_T\left(\lambda\right)=\lambda^2+2$

whose roots are $-\sqrt{2}i$ and $\sqrt{2}i$. \

Thus by Theorem 9.3.9 we have that the general solution to the differential equation $\frac{d^2x}{dt^2}+2x=0$ is

$x(t)=c_1\cos(\sqrt{2}t)+c_2e^t\sin(\sqrt{2}t)$

Let's look now for a particular solution to the equation $\frac{d^2x}{dt^2}+2x=\cos(t)$ of the form

$x_p(t)=A\cos(t)+B\sin(t)$

If $x_p(t)=A\cos(t)+B\sin(t)$ then $\frac{d^2x_p(t)}{dt^2}=-A\cos(t)-B\sin(t)$
which gives us that \

$\frac{d^2x_p(t)}{dt^2}+2x_p(t)=-A\cos(t)-B\sin(t)+2\left(A\cos(t)+B\sin(t)\right)=$

$=-A\cos(t)-B\sin(t)+2A\cos(t)+2B\sin(t)=A\cos(t)+B\sin(t)$

Thus in order for $x_p(t)$ to be a particular solution to the equation $\frac{d^2x}{dt^2}+2x=\cos(t)$

we must have $A\cos(t)+B\sin(t)=\cos(t)$. \

Equation $A\cos(t)+B\sin(t)=\cos(t)$ gives us that $A=1$ and $B=0$. Thus

$x_p(t)=\cos(t)$

Hence the general solution to the differential equation is \

$x(t)=c_1\cos(\sqrt{2}t)+c_2e^t\sin(\sqrt{2}t)+\cos(t)$

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