Solve the subsequent IVP.

y′′ + 11y′ + 24 y = 0

y (0) =0

 y′ (0)=-7

 Solution

The characteristic equation is as

r² +11r + 24 = 0

( r + 8) ( r + 3) = 0

 Its roots are r₁ = - 8 and r₂ = -3 and then the general solution and its derivative is.

y (t ) = ce_-8t  + c e^-3t

 y′ (t ) = -8c₁e ^-8t - 3c₂e ^-3t

 Here, plug in the initial conditions to find the subsequent system of equations.

0 = y (0) = c1 + c2

-7 = y′ (0) = -8c₁ - 3c₂

Solving this system provides c₁ = 7and c₂ = - 7. The actual solution to the differential equation is so y (t ) = (7/5) e^-8t  - (7/5) e^-3t