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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 7 y^{^{''}} + 8 y^{^{'}} - 16 y = - 2 e^{x} - 2

y^{^{''}} + 3 y^{^{'}} + 2 y = 42 \cdot e^{5} x

x^{^{'''}} + 3 x^{^{''}} + x^{^{'}} + 3 x = t^{2}

y^{^{''}} + 3 y^{^{'}} + 15.25 y = e^{t}

y^{^{''}} - y^{^{'}} + \frac{y}{4} = 4 + e^{\frac{x}{2}}

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