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• Notes

Use the Laplace transform to solve the initial value problem
$y^{\prime \prime}(t)+y(t)=\sin (t), \quad y(0)=y^{\prime}(0)=0$

$\mathcal{L}[y(t)]=y(s),$ Take $L$

$L\left[y^{\prime \prime}(t)\right]+L[y(t)]=L[\sin t]$

$s^{2} y(s)-s y(0)-y^{\prime}(0)+y(s)=\frac{1}{s^{2}+1}$

$y(s)\left[s^{2}+1\right]=\frac{1}{s^{2}+1} \rightarrow y(s)=\frac{1}{\left(s^{2}+1\right)^{2}}$

$y(t)=L^{-1}\left[\frac{1}{\left(s^{2}+1\right)^{2}}\right]=\int_{0}^{t} \sin \beta \sin (t-\beta) d \beta$

$=\int_{0}^{t} \frac{1}{2}(\cos (\beta-t+\beta)-\cos (\beta+t-\beta)) d \beta$

$=\int_{0}^{t} \frac{1}{2}(\cos (2 \beta-t)-\cos (t) d \beta$

$=\frac{1}{2}\left[\left(\frac{1}{2}(\sin (2 \beta-t)-\cos (t) \beta)\right]_{0}^{t}\right.$

$=\frac{1}{2}\left[\frac{1}{2} \sin (t)-t \cos (t)-\left(-\frac{1}{2} \sin t-0\right)\right]$

$=\frac{1}{2}\left[\frac{1}{2} \sin (t)-t \cos (t)+\frac{1}{2} \sin (t)\right]$

$=\frac{1}{2}[\sin (t)-t \cos (t)]$

Solve the integral equation: $y(t)=t+\int_{0}^{t} y(\beta) \sin 2(t-\beta) \mathrm{d} \beta$

Take $L \rightarrow L[y(t)]=L[t]+L y(t) \cdot L[\sin 2 t]$

$y(s)=\frac{1}{s^{2}}+y(s) \cdot \frac{2}{s^{2}+4} \quad * s^{2}+4$

$\left(s^{2}+4\right) y(s)=\frac{s^{2}+4}{s^{2}}+2 y(s)$

$\left(s^{2}+4\right) y(s)-2 y(s)=1+\frac{4}{s^{2}}$

$y(s)\left[s^{2}+4-2\right]=1+\frac{4}{s^{2}}$

$y(s)\left[s^{2}+2\right]=1+\frac{4}{s^{2}} \quad \div s^{2}+2$

$y(s)=\frac{1}{s^{2}+2}+\frac{4}{s^{2}\left(s^{2}+2\right)}$

$=\frac{1}{s^{2}+2}+4\left(\frac{1 / 2}{s^{2}}+\frac{-1 / 2}{s^{2}+2}\right)$

$y(t)=L^{-1}[y(s)]$

$y(t)=\frac{1}{\sqrt{2}} \sin (\sqrt{2} t)+2 t-\frac{2 \cdot 1}{\sqrt{2}} \sin \sqrt{2} t$