Solve the given initial-value problem.d2xdt2=ω2x=F0cos⁡(γt), x(0)=0, x′(0)=0

Find the homogeneous solution of the given IVP as follows.
The given IVP is as follows,
${\displaystyle \frac{d^{2}x}{{dt}^2}={\omega }^{2}x=F_0\cos \left(\gamma t\right),x\left(0\right)=0,x^{\prime }\left(0\right)=0}$ and ${\displaystyle \omega \ne \gamma }$
${\displaystyle \Rightarrow x{}^{″}+{\omega }^{2}x=F_0\cos \left(\gamma t\right),x\left(0\right)=x^{\prime }\left(0\right)=0}$ and ${\displaystyle \omega \ne \gamma }$
Consider the homogeneous part ${\displaystyle x{}^{″}+{\omega }^{2}x=0}$
The characterice equation is ${\displaystyle m^2+{\omega }^2=0}$
${\displaystyle \Rightarrow m^2+{\omega }^2=0}$
${\displaystyle \Rightarrow =-\omega }$
${\displaystyle \Rightarrow =-\omega i,\omega i}$
The homogeneous solution is ${\displaystyle x_h=C_1\cos \left(\omega t\right)+C_2\sin \left(\omega t\right)}$
Find the particular solution of the given IVP as follows.
The IVP is ${\displaystyle \Rightarrow x{}^{″}+{\omega }^{2}x=F_0\cos \left(\gamma t\right),x\left(0\right)=x^{\prime }\left(0\right)=0}$ and ${\displaystyle \omega \ne \gamma }$
Let the particular solution be of the form, ${\displaystyle x_p=A\cos \left(\gamma t\right)}$
${\displaystyle x^{\prime }=-\gamma A\sin \left(\gamma t\right)}$
${\displaystyle x{}^{″}=-{\gamma }^2\cos \left(\gamma t\right)}$
${\displaystyle \Rightarrow -{\gamma }^2\cos \left(\gamma t\right)+{\omega }^{2}A\cos \left(\gamma t\right)=F_0\cos \left(\gamma t\right)}$
${\displaystyle \Rightarrow A{\omega }^2-{\gamma }^2=F_0}$
${\displaystyle \Rightarrow A=\frac{F_0+{\gamma }^2}{{\omega }^2},\omega \ne 0}$
The particular solution is ${\displaystyle x_p=\frac{F_0+{\gamma }^2}{{\omega }^2}\cos \left(\gamma t\right)}$
The general solution is of the given IVP is as follows.
The general solution is
${\displaystyle x\left(t\right)=x_h+x_p}$
${\displaystyle =C_1\cos \left(\omega t\right)+C_2\sin \left(\omega t\right)+\frac{F_0+{\gamma }^2}{{\omega }^2}\cos \left(\gamma t\right)}$

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