What is the solution for q^{''}+20q^'+100q=24,q(0)=0,q^'(0)=0 ?

The solution for q^{''}+20q^'+100q=24,q(0)=0,q^'(0)=0 is q(t)=-6/25 e^{-10t}-(12e^{-10t}t)/5+6/25

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q^{''}+20q^'+100q=24,q(0)=0,q^'(0)=0

q^{^{''}} + 20 q^{^{'}} + 100 q = 24, q (0) = 0, q^{^{'}} (0) = 0

q (t) = - \frac{6}{25} e^{- 10 t} - \frac{12 e ^{- 10 t} t}{5} + \frac{6}{25}

Solution steps

q^{''} (t) + 20 q^{'} (t) + 100 q (t) = 24

Solve linear ODE:

q (t) = - \frac{6}{25} e^{- 10 t} - \frac{12 e ^{- 10 t} t}{5} + \frac{6}{25}

q (t) = - \frac{6}{25} e^{- 10 t} - \frac{12 e ^{- 10 t} t}{5} + \frac{6}{25}

What is the solution for q^{''}+20q^'+100q=24,q(0)=0,q^'(0)=0 ?
The solution for q^{''}+20q^'+100q=24,q(0)=0,q^'(0)=0 is q(t)=-6/25 e^{-10t}-(12e^{-10t}t)/5+6/25