What is the solution for y^{''}-3y^'-4y=6 ?

The solution for y^{''}-3y^'-4y=6 is y=c_{1}e^{4t}+c_{2}e^{-t}-3/2

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y^{''}-3y^'-4y=6

y^{^{''}} - 3 y^{^{'}} - 4 y = 6

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

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 6

Solve linear ODE:

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

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

3 y^{^{''}} + 192.08 y = 48.02 cos (8 t)

y^{^{''}} - 2 y^{^{'}} - 24 y = 16 - (x + 2) e^{4 x}

y^{^{''}} - 7 y^{^{'}} + 12 y = x

3 \frac{d ^{2} x}{d t ^{2}} + 10 x = 2 sin (3 t)

y^{^{''}} + y = 6 e^{- 8 x}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}-3y^'-4y=6 ?
The solution for y^{''}-3y^'-4y=6 is y=c_{1}e^{4t}+c_{2}e^{-t}-3/2