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Solve the initial value problem
$$t y^{\prime \prime}(t)+2 y^{\prime}(t)+t y(t)=0 \quad$$ if $$y(0)=1, y^{\prime}(0)=0$$

Let $$\mathcal {L}\{y(t)\}=y(s)$$

$$\mathcal {L} t y^{\prime \prime}(t)+ \mathcal {L} 2 y^{\prime}(t)+\mathcal {L} t y(t)=0$$

$$-\left[s^{2} y(s)-s y(0)-y^{\prime}(0)\right]^{\prime}+2[s y(s)-y(0)]-y^{\prime}(s)=0$$

$$-\left[2 s y(s)+s^{2} y^{\prime}(s)-y(0)-0\right]+2 s y(s)-2 y(0)-y^{\prime}(s)=0$$

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

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

$$\mathcal{L}\{-t y(t)\}=\frac{-1}{s^{2}+1} $$

$$\mathcal {L}^{-1} \mathcal{L}\{-t y(t)\}=\mathcal {L}^{-1}\frac{-1}{s^{2}+1} $$

$$t y(t)=\mathcal{L}^{-1}\left\{\frac{-1}{s^{2}+1}\right\} \longrightarrow$$

$$y(t)=\frac{\sin (t)}{t} $$

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