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• Notes

Solve:
$x_{1}^{\prime \prime}+10 x_{1}-4 x_2=0$
$-4 x_{1}+x_{2}^{\prime \prime}+4 x_{2}=0$
Where $x_{1}(0)=0, x_{1}^{\prime}(0)=1, x_{2}(0)=0, x_{2}^{\prime}(0)=-1$

Applying Laplace Trasform :

$\mathcal {L}\left\{x_{1}^{\prime \prime}\right\}+\mathcal {L}\left\{ 10 x_{1}\right\}+\mathcal {L}\left\{-4 x_{2}\right\}=0$

$s^{2} x_{1}(s)-s x_{1}(0)-x_{1}^{\prime}(0)+10 x_{1}(s)-4 x_{2}(s)=0$

$\mathcal{L}\left\{-4 x_{1}\right\}+\mathcal{L}\left\{x_{2}^{\prime \prime}\right\}+\mathcal{L}\left\{4 x_{2}\right\}=0$

$-4 X_{1}(s)+S^{2} X_{2}(S)-S X_{2}(0)-X_{2}^{\prime}(0)+4 X_{2} (S)=0$

$\rightarrow\left[s^{2}+10\right] x_{1}(s)-4 x_{2}(s)=1 \quad * s^{2}+4$

$\rightarrow-4 x_{1}(s)+\left[s^{2}+4\right] x_{2}(s)=-1 \quad * 4$

$\left[\left(s^{2}+10\right)\left(s^{2}+4\right)\right] x_1(s)-4\left(s^{2}+4\right) x_{2}(s)=s^{2}+4 \cdots (1)$

$-16 x_{1}(s)+4\left(s^{2}+4\right) x_{2}(s)=-4 \cdots (2)$

$\left[\left(s^{2}+10\right)\left(s^{2}+4\right)-16\right] x_{1}(s)=\left(s^{2}+4\right)-4$

$\rightarrow x_{1}(s)=\frac{s^{2}}{\left(s^{2}+10\right)\left(s^{2}+4\right)-16}$

$x_{1}(s)=\frac{s^{2}}{s^{4}+4 s^{2}+10 s^{2}+40-16}=\frac{s^{2}}{s^{4}+14 s^{2}+24}$

By using Partial fractions:

$x_{1}(s)=\frac{-1 / 5}{s^{2}+2}+\frac{6 / 5}{s^{2}+12}$

$x_{2}(s)=\frac{-s^{2}+6}{\left(s^{2}+2\right)\left(s^{2}+12\right)}=\frac{-2 / 5}{s^{2}+2}-\frac{3 / 5}{s^{2}+12}$

Apply laplaes inverse in $x_{1}(s) \& \ x_{2}(s) :$

$x_{1}(t)=\frac{-\sqrt{2}}{10} \sin \sqrt{2} t+\frac{\sqrt{3}}{5} \sin 2 \sqrt{3} t$

$x_{2}(t)=\frac{-\sqrt{2}}{5} \sin \sqrt{2} t-\frac{\sqrt{3}}{10} \sin 2 \sqrt{3} t$