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The problem can be solved by applying Duhamel-theorem.
$                  y(t)=f(t)h(0)+\int_{0}^{t}f(\tau )\frac{dh(t-\tau)}{d\tau}d\tau, or let’s apply the other formula:

$                  y(t)=f(0)h(t)+\int_{0}^{t} \frac{d f(\tau) }{d\tau} h(t-\tau)d\tau

The latter formula will be used in order to determine the output response:
$                  y(t)=-(F_0\text{sin}\varphi)A(e^{s_1t}- e^{s_2t})+\int_{0}^{t} \frac{d F_0\text{sin}(\omega \tau+\varphi) }{d\tau} \left [ -A(e^{s_1(t-\tau)}- e^{s_2(t-\tau)}) \right ] d\tau

$                  y(t)=-(AF_0\text{sin}\varphi)(e^{s_1t}- e^{s_2t})-\int_{0}^{t} \frac{d F_0\text{sin}(\omega \tau+\varphi) }{d\tau} A \left [ (e^{s_1(t-\tau)}- e^{s_2(t-\tau)}) \right ] d\tau

$                  y(t)=-AF_0\text{sin}\varphi(e^{s_1t}- e^{s_2t})+\int_{0}^{t} AF_0\omega \text{cos}(\omega \tau+\varphi) \left [ (e^{s_1(t-\tau)}- e^{s_2(t-\tau)}) \right ] d\tau

$                  y(t)=-AF_0 (e^{s_1t}- e^{s_2t}) \text{sin}\varphi + AF_0\omega \int_{0}^{t} \text{cos}(\omega \tau+\varphi) \left [ (e^{s_1(t-\tau)}- e^{s_2(t-\tau)}) \right ] d\tau

Let’s execute the following integral $               \int_{0}^{t} e^{s_1(t-\tau)}\text{cos}(\omega \tau+\varphi)   d\tau .

Partial integral method will be used.
$                  u= \text{cos}(\omega \tau+\varphi) \quad {u}'=-\omega \text{sin}(\omega \tau+\varphi) 
\quad \quad {v}'= e^{s_1(t-\tau)} \quad v=\frac{-1}{s_1} e^{s_1(t-\tau)}

$              \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)   d\tau  = \left [ -\frac{1}{s_1} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi) \right ]_0^t-\frac{\omega}{s_1}\int_{0}^{t} e^{s_1(t-\tau)} \text{sin}(\omega \tau+\varphi)  d\tau

$          \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)   d\tau  = \frac{1}{s_1} \left [e^{s_1t} \text{cos}(\varphi)-\text{cos}(\omega t+\varphi) \right ] -\frac{\omega}{s_1}\int_{0}^{t} e^{s_1(t-\tau)} \text{sin}(\omega \tau+\varphi)  d\tau

Let’s apply again the partial integral method.
$                  u= \text{sin}(\omega \tau+\varphi) \quad u'=\omega \text{cos}(\omega \tau+\varphi) 
\quad \quad v'= e^{s_1(t-\tau} \quad v=\frac{-1}{s_1} e^{s_1(t-\tau)}

$            \frac{\omega}{s_1}\int_{0}^{t} e^{s_1(t-\tau)} \text{sin}(\omega \tau+\varphi)  d\tau  =\frac{\omega}{s_1} \left [ \left [-\frac{1}{s_1} e^{s_1(t-\tau)}) \text{sin}(\omega \tau+\varphi)  \right ]_0^t+\frac{\omega}{s_1} \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau  \right ]

$          \frac{\omega}{s_1}\int_{0}^{t} e^{s_1(t-\tau)} \text{sin}(\omega \tau+\varphi)  d\tau  =\frac{\omega}{s_1} \left [\frac{1}{s_1} e^{s_1 t} \text{sin}(\varphi) - \frac{ \text{sin}(\omega t+\varphi)  }{s_1}  \right ]+\frac{\omega^2}{s_1^2}\int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau

Now the $        \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau integral element can be expressed as follow:
$           \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau  = \frac{1}{s_1} \left [e^{s_{1}t} \text{cos}(\varphi)-\text{cos}(\omega t+\varphi) \right ] - \frac{\omega}{s_1^2} \left [ e^{s_{1}t} \text{sin}( \varphi) - \text{sin}( \omega t + \varphi) \right ] - \frac{\omega^2}{s_1^2}\int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau

The unknown integral element:

$        (s_1^2+\omega^2) \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau  = \left [ -s_1\text{cos}(\omega t +\varphi)+\omega \text{sin}(\omega t + \varphi) + e^{s_1t}( s_1\text{cos}(\varphi)-\omega \text{sin}(\varphi)) \right ]

$                  \int_{0}^{t} e^{s_1(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau  = \frac{1}{ s_1^2+\omega^2}\left [ -s_1\text{cos}(\omega t +\varphi)+\omega \text{sin}(\omega t + \varphi) + e^{s_1t}( s_1\text{cos}(\varphi)-\omega \text{sin}(\varphi)) \right ]

Completely same way can be written the other integral element:

$                  \int_{0}^{t} e^{s_2(t-\tau)} \text{cos}(\omega \tau+\varphi)  d\tau  = \frac{1}{ s_2^2+\omega^2}\left [ -s_2\text{cos}(\omega t +\varphi)+\omega \text{sin}(\omega t + \varphi) + e^{s_2t}( s_2\text{cos}(\varphi)-\omega \text{sin}(\varphi)) \right ]

After substituting both integral elements, the response will be the following!
y(t)=AF_0(e^{s_2t}- e^{s_1t}) \text{sin}\varphi + F_{0}A\omega \left [\frac{1}{ s_1^2+\omega^2}\left [ -s_1\text{cos}(\omega t +\varphi)+\omega \text{sin}(\omega t + \varphi) + e^{s_1t}( s_1\text{cos}(\varphi)-\omega \text{sin}(\varphi)) \right ]\right]  -

$ - F_{0}A\omega \left [ \frac{1}{ s_2^2+\omega^2}\left [ -s_2\text{cos}(\omega t +\varphi)+\omega \text{sin}(\omega t + \varphi) + e^{s_2t}( s_2\text{cos}(\varphi)-\omega \text{sin}(\varphi)) \right ] \right]

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