Question

como resolver el siguiente te problema con valores iniciales.

Answer 1

$\dfrac{d^2x}{dt^2}+6\dfrac{dx}{dt}+10x=25\cos 4t\\ \\ \\ \mathcal{L}\left\{\dfrac{d^2x}{dt^2}+6\dfrac{dx}{dt}+10x\right\}=\mathcal{L}\left\{25\cos 4t\right\}\\ \\ \\ \mathcal{L}\left\{\dfrac{d^2x}{dt^2}\right\}+6\mathcal{L}\left\{\dfrac{dx}{dt}\right\}+10\mathcal{L}\left\{x\right\}=25\mathcal{L}\left\{\cos 4t\right\}$


$\left[s^2\mathcal{L}\left\{x\right\}-sx(0^-)-x'(0^-)\right]+6\left[s\mathcal{L}\left\{x\right\}-x(0^-)\right]+10\mathcal{L}\left\{x\right\}=\dfrac{25s}{s^2+4^2}\\ \\ \\ (s^2+6s+10)\mathcal{L}\{x\}-\dfrac{s}{2}-3=\dfrac{25s}{s^2+16}\\ \\ \\ \mathcal{L}\{x\}=\dfrac{\dfrac{25s}{s^2+16}+\dfrac{s}{2}+3}{s^2+6s+10}$


$\mathcal{L}\{x\}=\dfrac{\frac{38}{51}s}{s^2+6s+10}+\dfrac{\frac{28}{51}}{s^2+6s+10}-\dfrac{\frac{25}{102}s}{s^2+16}+\dfrac{\frac{200}{51}}{s^2+16}\\ \\ \\ x(t)=\dfrac{38}{51}\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+6s+10}\right\}+\dfrac{28}{51}\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+6s+10}\right\}-\cdots\\ \\ -\dfrac{25}{102}\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+16}\right\}+\dfrac{200}{51}\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+16}\right\}$


$x(t)=\dfrac{38}{51}\mathcal{L}^{-1}\left\{\dfrac{s}{(s+3)^2+1}\right\}+\dfrac{28}{51}\mathcal{L}^{-1}\left\{\dfrac{1}{(s+3)^2+1}\right\}-\cdots\\ \\ -\dfrac{25}{102}\cos(4t)+\dfrac{50}{51}\sin(4t)$


$x(t)=\dfrac{38}{51}\mathcal{L}^{-1}\left\{\dfrac{s+3}{(s+3)^2+1}\right\}-\dfrac{86}{51}\mathcal{L}^{-1}\left\{\dfrac{1}{(s+3)^2+1}\right\}-\cdots\\ \\ -\dfrac{25}{102}\cos(4t)+\dfrac{50}{51}\sin(4t)\\ \\ \\ \texttt{Recordando la propiedad de traslaci\'on}\\ \\ \boxed{\mathcal{L}^{-1}\left\{F(s-\omega)\right\}=e^{\omega t}\mathcal{L}^{-1}\left\{F(s)\right\}}$


$x(t)=\dfrac{38e^{-3t}}{51}\mathcal{L}^{-1}\left\{\dfrac{s}{s^2+1}\right\}-\dfrac{86e^{-3t}}{51}\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+1}\right\}-\cdots\\\\ \cdots-\dfrac{25}{102}\cos(4t)+\dfrac{50}{51}\sin(4t)\\ \\ \\ \\ \boxed{x(t)=\dfrac{38e^{-3t}}{51}\cos t-\dfrac{86e^{-3t}}{51}\sin t-\dfrac{25}{102}\cos(4t)+\dfrac{50}{51}\sin(4t)}$