$\displaystyle \frac{dy}{dt}+2y+\int_0^tydt=\text{sin} \ t\ \[3ex] Let \ \ L(y) =\bar{y}\[2ex] Taking \ Laplace \ transform \ on \ both \ sides,\[2ex] \displaystyle L(y') + 2L(y) + L\left[\int_0^tydt\right]= L (\sin t)\[2ex] L(y') = sL(y)- y(0) = s\bar{y}-1\[3ex] \begin{align} L\left[\int_0^tydt\right] &= \frac{1}{s}L(y) \[2ex] &= \frac{1}{s}\bar{\ y}\[2ex] L(\sin t) &= \frac{1}{s^2+1} \end{align}$ $\ \therefore{}s\bar{y}-1+2\bar{y}+\ \frac{1}{s}\bar{y}=\dfrac{1}{s^2+1}\[2ex] \begin{align} \therefore{}\left(s+2+\frac{1}{s}\right)\bar{\ y}&=\ \frac{1}{s^2+1}+1\[2ex] &=\ \frac{s^2+1+1}{s^2+1}\[2ex] \therefore{}\left(\frac{s^2+2s+1}{s}\right)\bar{y} &=\frac{s^2+2}{s^2+1}\[2ex] \therefore{} \bar{y}&=\frac{s\left(s^2+2\right)}{{(s+1)}^2\left(s^2+1\right)}\[2ex] \end{align}$ ***By partial fraction,*** $Let,\[1ex] \bar{y}=\ \dfrac{s\left(s^2+2\right)}{{(s+1)}^2\left(s^2+1\right)} \[2ex] \displaystyle \bar{y}=\ \frac{a}{s+1}+\ \frac{b}{{(s+1)}^2}+\frac{cs+d}{s^2+1}\[3ex] \therefore{} \ s\left(s^2+2\right)= a(s + 1)( s^2+1) + b(s^2+1) + \left(cs+d\right){(s+1)}^2\[2ex] $ ***Putting s = -1*** $-3\ = 2b\[2ex]\[2ex] \therefore{} b = -\dfrac{3}{2}$ ***Putting s = 0*** $0 = a + b + d \[1ex] but, \ \ b = -\dfrac{3}{2} ,\[2ex] a + d = \dfrac{3}{2}\[2ex] s^3+2s=a\left[s^3+s+s^2+1\right]+b\left[s^2+1\right]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +cs\left[s^2+2s+1\right]+d[s^2+2s+1]$ Equating the coefficient of s2 and s3 $0 = a+ b + 2c + d ...(1) \[2ex] 1 = a + c...(2)\[3ex] Putting \ in (1) \ b = -\dfrac{3}{2} ,\ a + d = \dfrac{3}{2}\[2ex] \ \ \ 2c = 0 \[2ex] ∴c=0\[3ex] Putting \ in (2)\[2ex] a = 1\[1ex] but \ a + d = \dfrac{3}{2}\[2ex] d = \dfrac{1}{2}$ $\displaystyle \therefore{}\ \bar{y}=\ \frac{1}{s+1}-\frac{3}{2}\ \frac{1}{{(s+1)}^2}+\frac{1}{2}\ \frac{1}{s^2+1}\[3ex] \displaystyle \therefore{} y = L^{-1}\frac{1}{s+1}-\frac{3}{2}\ L^{-1}\frac{1}{{\left(s+1\right)}^2}+\frac{1}{2}L^{-1}\ \frac{1}{s^2+1}\[3ex] \displaystyle \therefore{} y = {e^{-t}L}^{-1}\frac{1}{s}-\frac{3}{2}\ {e^{-t}L}^{-1}\frac{1}{{\left(s\right)}^2}+\frac{1}{2}L^{-1}\ \frac{1}{s^2+1}\[3ex] \displaystyle \therefore{} y = e^{-t}-\ \frac{3}{2}e^{-t}.t + \frac{1}{2} \sin t\[2ex] $