Characteristic equation:
[tex]r^2+4r-5=(r+5)(r-1)=0\implies r=-5,r=1[/tex]
[tex]\implies y_c=C_1e^{-5x}+C_2e^x[/tex]
Given that [tex]y(0)=0[/tex] and [tex]y\left(\dfrac\pi2\right)=1[/tex], you have
[tex]\begin{cases}0=C_1+C_2\\1=C_1e^{-5\pi/2}+C_2e^{\pi/2}\end{cases}\implies C_1=\dfrac{e^{5\pi/2}}{1-e^{3\pi}},C_2=-\dfrac{e^{5\pi/2}}{1-e^{3\pi}}[/tex]
so that the particular solution is
[tex]y=\dfrac{e^{5\pi/2}}{1-e^{3\pi}}(e^{-5x}-e^x)[/tex]
