Question Number 219491 by OmoloyeMichael last updated on 26/Apr/25
$$\boldsymbol{{if}}\:\boldsymbol{{x}}''−\mathrm{2}\boldsymbol{{x}}'+\mathrm{10}\boldsymbol{{x}}=\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{t}}} ,\:\boldsymbol{{at}}\:\boldsymbol{{t}}=\mathrm{0},\boldsymbol{{x}}=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{x}}'=\mathrm{1} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)\:\boldsymbol{{using}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}} \\ $$
Answered by mahdipoor last updated on 26/Apr/25
![Laplace ⇒ (Xs^2 −x(0)s−x^′ (0))−2(Xs−x(0))+10X=(1/(s−2)) ⇒X(s^2 −2s+10)−1=(1/(s−2)) ⇒X=((s−1)/((s−2)(s^2 −2s+10)))= ((0.1)/(s−2))+((−0.1(s−1))/((s−1)^2 +3^2 ))+((0.3×(3))/((s−1)^2 +3^2 )) Laplace^(−1) ⇒ 0.1[e^(2t) −e^t .cos(3t)+3e^t .sin(3t)]](https://www.tinkutara.com/question/Q219493.png)
$${Laplace}\:\Rightarrow \\ $$$$\left({Xs}^{\mathrm{2}} −{x}\left(\mathrm{0}\right){s}−{x}^{'} \left(\mathrm{0}\right)\right)−\mathrm{2}\left({Xs}−{x}\left(\mathrm{0}\right)\right)+\mathrm{10}{X}=\frac{\mathrm{1}}{{s}−\mathrm{2}} \\ $$$$\Rightarrow{X}\left({s}^{\mathrm{2}} −\mathrm{2}{s}+\mathrm{10}\right)−\mathrm{1}=\frac{\mathrm{1}}{{s}−\mathrm{2}} \\ $$$$\Rightarrow{X}=\frac{{s}−\mathrm{1}}{\left({s}−\mathrm{2}\right)\left({s}^{\mathrm{2}} −\mathrm{2}{s}+\mathrm{10}\right)}= \\ $$$$\frac{\mathrm{0}.\mathrm{1}}{{s}−\mathrm{2}}+\frac{−\mathrm{0}.\mathrm{1}\left({s}−\mathrm{1}\right)}{\left({s}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{0}.\mathrm{3}×\left(\mathrm{3}\right)}{\left({s}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} } \\ $$$$ \\ $$$${Laplace}^{−\mathrm{1}} \Rightarrow \\ $$$$\mathrm{0}.\mathrm{1}\left[{e}^{\mathrm{2}{t}} −{e}^{{t}} .{cos}\left(\mathrm{3}{t}\right)+\mathrm{3}{e}^{{t}} .{sin}\left(\mathrm{3}{t}\right)\right] \\ $$
Commented by OmoloyeMichael last updated on 26/Apr/25
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$