What is the solution for y^{''}-2y^'-8y=4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-2y^'-8y=4,y(0)=0,y^'(0)=0 is y=(-1/3+1/2)e^{4t}+1/3 e^{-2t}-1/2

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y^{''}-2y^'-8y=4,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} - 8 y = 4, y (0) = 0, y^{^{'}} (0) = 0

y = (- \frac{1}{3} + \frac{1}{2}) e^{4 t} + \frac{1}{3} e^{- 2 t} - \frac{1}{2}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 8 y = 4

Solve linear ODE:

y = (- \frac{1}{3} + \frac{1}{2}) e^{4 t} + \frac{1}{3} e^{- 2 t} - \frac{1}{2}

y = (- \frac{1}{3} + \frac{1}{2}) e^{4 t} + \frac{1}{3} e^{- 2 t} - \frac{1}{2}

\frac{d}{d x} (3 e^{x})

\int e^{\frac{x}{a}} d x

\int x 6 x - x^{2} d x

\int \frac{1}{x x + 9} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = 0

What is the solution for y^{''}-2y^'-8y=4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-2y^'-8y=4,y(0)=0,y^'(0)=0 is y=(-1/3+1/2)e^{4t}+1/3 e^{-2t}-1/2