What is the solution for y^{''}-3y^'-4y=6e^{2t} ?

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

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y^{''}-3y^'-4y=6e^{2t}

y^{^{''}} - 3 y^{^{'}} - 4 y = 6 e^{2 t}

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

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = 6 e^{2 t}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (e^{- (x^{2} + y^{2} + z^{2})})

x \to 0 + lim (\frac{3 x e ^{2 x}}{sin ( 6 x )})

y^{^{'}} = 0.0015 y (400 - y)

x \to + 0 lim ((1 - 2 x)^{\frac{3}{x}})

\frac{d}{d x} (b x^{2} (x - 4))

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