What is the solution for y^{''}+y^'-12y=4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y^'-12y=4,y(0)=0,y^'(0)=0 is y=(-1/7+1/3)e^{3t}+1/7 e^{-4t}-1/3

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y^{''}+y^'-12y=4,y(0)=0,y^'(0)=0

y^{^{''}} + y^{^{'}} - 12 y = 4, y (0) = 0, y^{^{'}} (0) = 0

y = (- \frac{1}{7} + \frac{1}{3}) e^{3 t} + \frac{1}{7} e^{- 4 t} - \frac{1}{3}

Solution steps

y^{''} (t) + y^{'} (t) - 12 y = 4

Solve linear ODE:

y = (- \frac{1}{7} + \frac{1}{3}) e^{3 t} + \frac{1}{7} e^{- 4 t} - \frac{1}{3}

y = (- \frac{1}{7} + \frac{1}{3}) e^{3 t} + \frac{1}{7} e^{- 4 t} - \frac{1}{3}

y^{^{''}} - 3 y^{^{'}} = x

y^{^{''}} + 12 y^{^{'}} + 100 y = 48 sin (10 t)

x^{^{''}} + 25 x = 102 cos (t)

y^{^{''}} - y^{^{'}} = 3 - 4 x

y^{^{''}} + 3 y + 2 y = cos (e^{x})

What is the solution for y^{''}+y^'-12y=4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y^'-12y=4,y(0)=0,y^'(0)=0 is y=(-1/7+1/3)e^{3t}+1/7 e^{-4t}-1/3