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}

x y^{^{'}} - y + 2 = 0

\int (2 sin (x) - 3 cos (x)) d x

20 t^{2}

x \to 17 lim (x^{2} - 3 x + 1)

\frac{1}{3} (x^{2} + 2)^{\frac{3}{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}