$$ay''+by'+cy=\frac{1}{1+{{e}^{-t}}}$$ Take the Laplace transform valid for t>0, $$a\left( -y{{'}_{0}}-s{{y}_{0}}+{{s}^{2}}Y \right)+b\left( sY-{{y}_{0}} \right)+cY=\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}$$ so $$Y\left( s \right)=\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a{{s}^{2}}+bs+c}+\frac{1}{a{{s}^{2}}+bs+c}\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}$$ Note: $$\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}}\int\limits_{0}^{\infty }{{{e}^{-\left( s+n \right)t}}dt}=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}}$$ And therefore upon inverting we have $$y\left( t \right)=\frac{1}{2\pi i}\int\limits_{{}}^{{}}{{}}\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a{{s}^{2}}+bs+c}{{e}^{st}}+\frac{1}{a{{s}^{2}}+bs+c}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}{{e}^{st}}}$$ Where the integral is the Bromwich contour. Let $a{{s}^{2}}+bs+c=a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)$ and let ${{\beta }_{\pm }}=\frac{-b\pm \sqrt{\Delta }}{2a}\Rightarrow a\left( {{\beta }_{+}}-{{\beta }_{-}} \right)=\sqrt{\Delta }\\$ where we have the discriminant $\Delta ={{b}^{2}}-4ac$. Then

$$y\left( t \right)=\frac{1}{2\pi i}\int\limits_{{}}^{{}}{{}}\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)}{{e}^{st}}+\frac{1}{a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}{{e}^{st}}}$$ Calculating residues we have $$y\left( t \right)=\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{+}}+n}} \right){{e}^{{{\beta }_{+}}t}}-\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{-}}+n}} \right){{e}^{{{\beta }_{-}}t}}+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$ It is then a reasonably simple matter to show that this solution satisfies the DE and the boundary conditions. For example, consider $$ay''+by'+cy=\frac{1}{\sqrt{\Delta }}\left\{ \begin{align} & \left( a{{\beta }^{2}}_{+}+b{{\beta }_{+}}+c \right)\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{+}}+n}} \right){{e}^{{{\beta }_{+}}t}} \\ & -\left( a{{\beta }^{2}}_{-}+b{{\beta }_{-}}+c \right)\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{-}}+n}} \right){{e}^{{{\beta }_{-}}t}} \\ \end{align} \right\}+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( a{{n}^{2}}-bn+c \right)}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$ Now the first two terms disappear because $\beta $ is a solution of the quadratic. We have then

$$ay''+by'+cy=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( a{{n}^{2}}-bn+c \right)}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$

Note $a{{s}^{2}}+bs+c=a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)$ hence $a{{n}^{2}}-bn+c=a\left( -n-{{\beta }_{+}} \right)\left( -n-{{\beta }_{-}} \right)=a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)$.

$$ay''+by'+cy=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}{{e}^{-nt}}}=\frac{1}{1+{{e}^{-t}}}$$

To check the boundary conditions again observe $a\left( {{\beta }_{+}}-{{\beta }_{-}} \right)=\sqrt{\Delta }$, but also $\left( {{\beta }_{+}}+{{\beta }_{-}} \right)=-b/a$. Therefore after some simplification

$$y\left( 0 \right)=\frac{1}{\sqrt{\Delta }}\left\{ a{{y}_{0}}\left( {{\beta }_{+}}-{{\beta }_{-}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( {{\beta }_{-}}-{{\beta }_{+}} \right)}{\left( {{\beta }_{+}}+n \right)\left( {{\beta }_{-}}+n \right)}}+\sqrt{\Delta }\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}} \right\}={{y}_{0}}$$

$$y'\left( 0 \right)=\frac{1}{\sqrt{\Delta }}\left\{ -\frac{b\sqrt{\Delta }}{a}{{y}_{0}}+\frac{\sqrt{\Delta }}{a}\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\frac{\left( {{\beta }_{+}}-{{\beta }_{-}} \right)n}{\left( {{\beta }_{+}}+n \right)\left( {{\beta }_{-}}+n \right)}}-\sqrt{\Delta }\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}n}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}} \right\}=y{{'}_{0}}$$. Now consider once more the solution y(t), and note it may be written in terms of some special functions $$\begin{align} & y\left( t \right)=\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\Phi \left( -1,1,{{\beta }_{+}} \right) \right){{e}^{{{\beta }_{+}}t}}-\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\Phi \left( -1,1,{{\beta }_{-}} \right) \right){{e}^{{{\beta }_{-}}t}} \\ & +\frac{1}{{{\beta }_{+}}{{\beta }_{-}}\sqrt{\Delta }}\left( {{\beta }_{+}}{}_{2}{{F}_{1}}\left( 1,{{\beta }_{-}};1+{{\beta }_{-}},{{e}^{-t}} \right)-{{\beta }_{-}}{}_{2}{{F}_{1}}\left( 1,{{\beta }_{+}};1+{{\beta }_{+}},{{e}^{-t}} \right) \right) \\ \end{align}$$ Where we have the Lerch transcendent and the hypergeometric function. This may be of interest perhaps, if you are considering certain types of zeros of the quadratic. For example note that $\Phi \left( -1,1,\beta \right)$ becomes undefined whenever $\beta <0$ which implies if we are considering two real solutions, then they must be positive which then places restrictions upon a,b,c etc. Perhaps also of interest is that it takes on specific forms depending upon the roots of the quadratic (as does the hypergeometric function).

$$ay''+by'+cy=\frac{1}{1+{{e}^{-t}}}$$ Take the Laplace transform valid for t>0, $$a\left( -y{{'}_{0}}-s{{y}_{0}}+{{s}^{2}}Y \right)+b\left( sY-{{y}_{0}} \right)+cY=\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}$$ so $$Y\left( s \right)=\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a{{s}^{2}}+bs+c}+\frac{1}{a{{s}^{2}}+bs+c}\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}$$ Note: $$\int\limits_{0}^{\infty }{\frac{{{e}^{-st}}}{1+{{e}^{-t}}}dt}=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}}\int\limits_{0}^{\infty }{{{e}^{-\left( s+n \right)t}}dt}=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}}$$ And therefore upon inverting we have $$y\left( t \right)=\frac{1}{2\pi i}\int\limits_{{}}^{{}}{{}}\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a{{s}^{2}}+bs+c}{{e}^{st}}+\frac{1}{a{{s}^{2}}+bs+c}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}{{e}^{st}}}$$ Where the integral is the Bromwich contour. Let $a{{s}^{2}}+bs+c=a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)$ and let ${{\beta }_{\pm }}=\frac{-b\pm \sqrt{\Delta }}{2a}\Rightarrow a\left( {{\beta }_{+}}-{{\beta }_{-}} \right)=\sqrt{\Delta }\\$ where we have the discriminant $\Delta ={{b}^{2}}-4ac$. Then

$$y\left( t \right)=\frac{1}{2\pi i}\int\limits_{{}}^{{}}{{}}\frac{a{{y}_{0}}s+\left( {{y}_{0}}b+ay{{'}_{0}} \right)}{a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)}{{e}^{st}}+\frac{1}{a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)}\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{s+n}{{e}^{st}}}$$ Calculating residues we have $$y\left( t \right)=\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{+}}+n}} \right){{e}^{{{\beta }_{+}}t}}-\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{-}}+n}} \right){{e}^{{{\beta }_{-}}t}}+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$ It is then a reasonably simple matter to show that this solution satisfies the DE and the boundary conditions. For example, consider $$ay''+by'+cy=\frac{1}{\sqrt{\Delta }}\left\{ \begin{align} & \left( a{{\beta }^{2}}_{+}+b{{\beta }_{+}}+c \right)\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{+}}+n}} \right){{e}^{{{\beta }_{+}}t}} \\ & -\left( a{{\beta }^{2}}_{-}+b{{\beta }_{-}}+c \right)\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{{{\beta }_{-}}+n}} \right){{e}^{{{\beta }_{-}}t}} \\ \end{align} \right\}+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( a{{n}^{2}}-bn+c \right)}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$ Now the first two terms disappear because $\beta $ is a solution of the quadratic. We have then

$$ay''+by'+cy=\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( a{{n}^{2}}-bn+c \right)}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}{{e}^{-nt}}}$$

Note $a{{s}^{2}}+bs+c=a\left( s-{{\beta }_{+}} \right)\left( s-{{\beta }_{-}} \right)$ hence $a{{n}^{2}}-bn+c=a\left( -n-{{\beta }_{+}} \right)\left( -n-{{\beta }_{-}} \right)=a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)$.

$$ay''+by'+cy=\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}{{e}^{-nt}}}=\frac{1}{1+{{e}^{-t}}}$$

To check the boundary conditions again observe $a\left( {{\beta }_{+}}-{{\beta }_{-}} \right)=\sqrt{\Delta }$, but also $\left( {{\beta }_{+}}+{{\beta }_{-}} \right)=-b/a$. Therefore after some simplification

$$y\left( 0 \right)=\frac{1}{\sqrt{\Delta }}\left\{ a{{y}_{0}}\left( {{\beta }_{+}}-{{\beta }_{-}} \right)+\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}\left( {{\beta }_{-}}-{{\beta }_{+}} \right)}{\left( {{\beta }_{+}}+n \right)\left( {{\beta }_{-}}+n \right)}}+\sqrt{\Delta }\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}} \right\}={{y}_{0}}$$

$$y'\left( 0 \right)=\frac{1}{\sqrt{\Delta }}\left\{ -\frac{b\sqrt{\Delta }}{a}{{y}_{0}}+\frac{\sqrt{\Delta }}{a}\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\sum\limits_{n=0}^{\infty }{{{\left( -1 \right)}^{n}}\frac{\left( {{\beta }_{+}}-{{\beta }_{-}} \right)n}{\left( {{\beta }_{+}}+n \right)\left( {{\beta }_{-}}+n \right)}}-\sqrt{\Delta }\sum\limits_{n=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}n}{a\left( n+{{\beta }_{+}} \right)\left( n+{{\beta }_{-}} \right)}} \right\}=y{{'}_{0}}$$. Now consider once more the solution y(t), and note it may be written in terms of some special functions $$\begin{align} & y\left( t \right)=\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{+}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\Phi \left( -1,1,{{\beta }_{+}} \right) \right){{e}^{{{\beta }_{+}}t}}-\frac{1}{\sqrt{\Delta }}\left( a{{y}_{0}}{{\beta }_{-}}+\left( {{y}_{0}}b+ay{{'}_{0}} \right)+\Phi \left( -1,1,{{\beta }_{-}} \right) \right){{e}^{{{\beta }_{-}}t}} \\ & +\frac{1}{{{\beta }_{+}}{{\beta }_{-}}\sqrt{\Delta }}\left( {{\beta }_{+}}{}_{2}{{F}_{1}}\left( 1,{{\beta }_{-}};1+{{\beta }_{-}},{{e}^{-t}} \right)-{{\beta }_{-}}{}_{2}{{F}_{1}}\left( 1,{{\beta }_{+}};1+{{\beta }_{+}},{{e}^{-t}} \right) \right) \\ \end{align}$$ Where we have the Lerch transcendent and the hypergeometric function. This may be of interest perhaps, if you are considering certain types of zeros of the quadratic. For example note that $\Phi \left( -1,1,\beta \right)$ becomes undefined whenever $\beta <0$ and an integer, which implies if we are considering two real integer solutions, then they must be positive which then places restrictions upon a,b,c etc. Perhaps also of interest is that it takes on specific forms depending upon the roots of the quadratic (as does the hypergeometric function).