What is the solution for y^{''}-6y^'+8y=-2cos(4t) ?

The solution for y^{''}-6y^'+8y=-2cos(4t) is y=c_{1}e^{4t}+c_{2}e^{2t}+3/40 sin(4t)+1/40 cos(4t)

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y^{''}-6y^'+8y=-2cos(4t)

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 cos (4 t)

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{3}{40} sin (4 t) + \frac{1}{40} cos (4 t)

Solution steps

y^{''} (t) - 6 y^{'} (t) + 8 y = - 2 cos (4 t)

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{3}{40} sin (4 t) + \frac{1}{40} cos (4 t)

y = c_{1} e^{4 t} + c_{2} e^{2 t} + \frac{3}{40} sin (4 t) + \frac{1}{40} cos (4 t)

\int_{0}^{2} 8 t d t

f (t) = t

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

\int \frac{12}{x ^{2} + 1} d x

What is the solution for y^{''}-6y^'+8y=-2cos(4t) ?
The solution for y^{''}-6y^'+8y=-2cos(4t) is y=c_{1}e^{4t}+c_{2}e^{2t}+3/40 sin(4t)+1/40 cos(4t)