What is the solution for y^{''}-6y^'+8y=-4e^{3t} ?

The solution for y^{''}-6y^'+8y=-4e^{3t} is y=c_{1}e^{4t}+c_{2}e^{2t}+4e^{3t}

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y^{''}-6y^'+8y=-4e^{3t}

y^{^{''}} - 6 y^{^{'}} + 8 y = - 4 e^{3 t}

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = - 4 e^{3 t}

Solve linear ODE:

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

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

\frac{\partial ^{2}}{\partial y ^{2}} (2 cos (x y^{2}))

f (x) = 4 x^{2} - 5 x, at x = - 3

\frac{\partial}{\partial x} (x^{2} ln (y - 1))

\int_{1}^{2} x (x - 1)^{4} d x

\int ln (t) \cdot t^{2} d t

What is the solution for y^{''}-6y^'+8y=-4e^{3t} ?
The solution for y^{''}-6y^'+8y=-4e^{3t} is y=c_{1}e^{4t}+c_{2}e^{2t}+4e^{3t}