solve $\displaystyle y'' - 3y' - 10 y = 0; \quad y = 1, y'=10 \;\textnormal{at}\; x = 0$

$\displaystyle y(x) = $

heres what i did:

$\displaystyle r^2-3r-10 = 0 $

$\displaystyle (r-2) (r-5) = 0 $

roots are : -2, 5

General solution: $\displaystyle y = C_1 e^{-2x}+C_2 e^{5x} $

$\displaystyle y' = -2C_1 e^{-2x}+5C_2 e^{5x} $

i need help applying the condition