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

The solution for y^{''}-4y^'-12y=0 is y=c_{1}e^{6t}+c_{2}e^{-2t}

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y^{''}-4y^'-12y=0

y^{^{''}} - 4 y^{^{'}} - 12 y = 0

y = c_{1} e^{6 t} + c_{2} e^{- 2 t}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 12 y = 0

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 2 t}

y = c_{1} e^{6 t} + c_{2} e^{- 2 t}

2 y

\frac{d y}{d x}, y = arccos (4 x^{2})

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

\int_{0}^{4} ((4 - x) x) d x

\frac{d y}{d x} = e^{4 x} - 3 y

What is the solution for y^{''}-4y^'-12y=0 ?
The solution for y^{''}-4y^'-12y=0 is y=c_{1}e^{6t}+c_{2}e^{-2t}