What is the solution for y^{''}-3y^'+2y=4cos(4t) ?

The solution for y^{''}-3y^'+2y=4cos(4t) is y=c_{1}e^{2t}+c_{2}e^t-12/85 sin(4t)-14/85 cos(4t)

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

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

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{12}{85} sin (4 t) - \frac{14}{85} cos (4 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{12}{85} sin (4 t) - \frac{14}{85} cos (4 t)

y = c_{1} e^{2 t} + c_{2} e^{t} - \frac{12}{85} sin (4 t) - \frac{14}{85} cos (4 t)

\int_{0}^{0.6} 250 x d x

\int - 2 sin (2 t) d t

\frac{d}{d x} y = (- 2. 6^{\frac{1}{2}})

ln (x^{61})

\frac{d}{d x} (\frac{5 x}{x ^{2} + 5})

What is the solution for y^{''}-3y^'+2y=4cos(4t) ?
The solution for y^{''}-3y^'+2y=4cos(4t) is y=c_{1}e^{2t}+c_{2}e^t-12/85 sin(4t)-14/85 cos(4t)