Elementary diff Eq question

[m_the0ry]

I took all my math too fast and I went and forgot it all.

Anyways I'm reviewing differential eq's and I need to solve

x''(t) + 9x'(t) + 14x(t) = -28

Where x'(0)= 0, x(0)=10. Now I remember how to do this for the form

ax''(t) + bx'(t) + cx(t) = 0

I know it has something to do with the characteristic polynomial and a general form but its all fuzzy in my head.

[_brady_]

Solve for the homogeneous equation:

r²+9r+14=0
(r+7)(r+2)=0

x1=C1exp(-7t)
x2=C2exp(-2t)

Solve for the inhomogeneous sol'n:

x(t)=x1+x2+X

The inhomogeneous part is of the form of a constant, so set X=A where A is a constant.

X'=X''=0 --> Plug into original diff eq

0+9*0+14*A=-28
A=-2

Therefore:

x(t)=C1exp(-7t)+C2exp(-2t)-2

Take first derivative, plug in BCs, and you get:

x(t)=(84/5)exp(-2t)-(24/5)exp(-7t)-2