I have the next differential equation: $R*C*y'(t)+y(t)=x(t)$ Where $x(t)$ is a piecewise funtion defined by: \begin{equation} \label{eq:aqui-le-mostramos-como-hacerle-la-llave-grande} x(t) = \left\{ \begin{array}{ll} 0 & \mathrm{if\ } t \le 0 \\ \alpha & \mathrm{if\ } 0 \le t \le \tau \\ -\frac{a}{\tau}+2\alpha & \mathrm{if\ } \tau \le t\le 2\tau\\ 0 & \mathrm{if\ } 2\tau \le t \end{array} \right. \end{equation} Where $R$,$C$,$\alpha$ and $\tau$ are constants with any value. And I know that the convolution is given by $\int_{-\infty}^{\infty}x(\lambda)h(t-\lambda)d\lambda$.
In this case $h(t)=\frac{1}{RC}e^{-\frac{t}{RC}}u(t)$.
Note: $h(t)$ was obtained by solving the next differential equation using the Laplace transforms: $RCh'(t)+h(t)=\delta(t)$
So using the convolution,I have two cases simpler than the others, when $t<0$ the function x(t) is equals to zero so in this case the value of the convolution is equals to zero.
And the other case is from $0\le t<\tau$ where I consider the next integral: \begin{equation*} y(t)=\int_0^t\alpha\frac{1}{RC}e^{-\frac{1}{RC}(t-\lambda)}d\lambda \end{equation*} What after do the integration, I got to this function of t: \begin{equation*} y(t)=\alpha(1-e^{-\frac{t}{RC}}) \end{equation*}
However I have problems to get the integral in the others intervals of the function.