Question

Find all real solutions o f the differential equations. d2xdt2+2x=cos(t)\frac{d^{2} x}{d t^{2}}+2 x=\cos (t)

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Consider the differential equation

d2xdt2+2x=cos(t)\frac{d^2x}{dt^2}+2x=\cos(t)

The characteristic polynomial of the operator T(x)=d2xdt2+2xT(x)=\frac{d^2x}{dt^2}+2x is

pT(λ)=λ2+2p_T\left(\lambda\right)=\lambda^2+2


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

Thus by Theorem 9.3.9 we have that the general solution to the differential equation d2xdt2+2x=0\frac{d^2x}{dt^2}+2x=0 is

x(t)=c1cos(2t)+c2etsin(2t)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 d2xdt2+2x=cos(t)\frac{d^2x}{dt^2}+2x=\cos(t) of the form

xp(t)=Acos(t)+Bsin(t)x_p(t)=A\cos(t)+B\sin(t)

If xp(t)=Acos(t)+Bsin(t)x_p(t)=A\cos(t)+B\sin(t) then d2xp(t)dt2=Acos(t)Bsin(t)\frac{d^2x_p(t)}{dt^2}=-A\cos(t)-B\sin(t)
which gives us that \

d2xp(t)dt2+2xp(t)=Acos(t)Bsin(t)+2(Acos(t)+Bsin(t))=\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)=

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

Thus in order for xp(t)x_p(t) to be a particular solution to the equation d2xdt2+2x=cos(t)\frac{d^2x}{dt^2}+2x=\cos(t)

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

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

xp(t)=cos(t)x_p(t)=\cos(t)

Hence the general solution to the differential equation is \

x(t)=c1cos(2t)+c2etsin(2t)+cos(t)x(t)=c_1\cos(\sqrt{2}t)+c_2e^t\sin(\sqrt{2}t)+\cos(t)

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