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

The solution for y^{''}-4y^'+3y=2cos(3t) is y=c_{1}e^{3t}+c_{2}e^t-2/15 sin(3t)-1/15 cos(3t)

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

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

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{2}{15} sin (3 t) - \frac{1}{15} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{2}{15} sin (3 t) - \frac{1}{15} cos (3 t)

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{2}{15} sin (3 t) - \frac{1}{15} cos (3 t)

f (x) = \frac{2 x - x ^{2}}{4 x ^{2} + 1}

\frac{d}{d x} (\frac{4}{7 x ^{3}})

\frac{d}{d x} ((6 x)^{6 x})

y^{^{'}} + y + 3 sin (x) = 0

y = x^{2} - 14, y = 7 - 4 x

What is the solution for y^{''}-4y^'+3y=2cos(3t) ?
The solution for y^{''}-4y^'+3y=2cos(3t) is y=c_{1}e^{3t}+c_{2}e^t-2/15 sin(3t)-1/15 cos(3t)