What is the solution for y^{''}-y^'-6y=-4e^{2t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}-y^'-6y=-4e^{2t},y(0)=0,y^'(0)=0 is y=(1/5-1)e^{3t}-1/5 e^{-2t}+e^{2t}

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y^{''}-y^'-6y=-4e^{2t},y(0)=0,y^'(0)=0

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{1}{5} - 1) e^{3 t} - \frac{1}{5} e^{- 2 t} + e^{2 t}

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = - 4 e^{2 t}

Solve linear ODE:

y = (\frac{1}{5} - 1) e^{3 t} - \frac{1}{5} e^{- 2 t} + e^{2 t}

y = (\frac{1}{5} - 1) e^{3 t} - \frac{1}{5} e^{- 2 t} + e^{2 t}

f (x) = \frac{2 - x ^{2}}{2 + x ^{2}}

f (x)

(- 5, 5), (5, - 5)

(ln (3))^{^{'}}

\frac{\partial}{\partial x} (z + 1 - x e^{y} cos (z))

What is the solution for y^{''}-y^'-6y=-4e^{2t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y^'-6y=-4e^{2t},y(0)=0,y^'(0)=0 is y=(1/5-1)e^{3t}-1/5 e^{-2t}+e^{2t}