What is the solution for y^{''}-2y^'-3y=4e^{3t} ?

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

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y^{''}-2y^'-3y=4e^{3t}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{3 t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 4 e^{3 t}

Solve linear ODE:

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

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

\int - 2 e^{c o s (x)} sin (x) d x

f (x) = 21

\int cos^{3} (4 x) sin^{2} (4 x) d x

\frac{\partial}{\partial x} (20 arctan (x))

(4 x^{4} - x + 3) (- x^{5} + 5)

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