What is the solution for y^{''}-3y^'-4y=12 ?

The solution for y^{''}-3y^'-4y=12 is y=c_{1}e^{4t}+c_{2}e^{-t}-3

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y^{''}-3y^'-4y=12

y^{^{''}} - 3 y^{^{'}} - 4 y = 12

y = c_{1} e^{4 t} + c_{2} e^{- t} - 3

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 12

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- t} - 3

y = c_{1} e^{4 t} + c_{2} e^{- t} - 3

\frac{\partial}{\partial y} (5 x^{2} + y^{2})

y = 0, y = 9 - x^{2}

\int cos^{2} (x) + 1 d x

y = - 4 lo g_{5} (x)

\frac{d}{d x} (e^{3 x} + e^{x + e^{- 2 x}})

What is the solution for y^{''}-3y^'-4y=12 ?
The solution for y^{''}-3y^'-4y=12 is y=c_{1}e^{4t}+c_{2}e^{-t}-3