 Solution

The constant term belongs on the forcing side: the equation is $y^{\prime \prime }-2y^{\prime }=x+1$. Set $v=y^{\prime }$ to obtain the first-order linear equation $v^{\prime }-2v=x+1$. A particular solution is $v_p=-x/2-3/4$, while $v_h=A{e}^{2x}$. Imposing $v(0)=0$ fixes $A=3/4$. Integrating and then using $y(0)=0$ givesDirect differentiation gives $y^{\prime }=3e^{2x}/4-x/2-3/4$ and $y^{\prime \prime }=3e^{2x}/2-1/2$, so $y^{\prime \prime }-2y^{\prime }-1=x$. Both initial values vanish. The nonzero exponential term is required by the initial derivative, even though the forcing is polynomial.