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)

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

\int \frac{4 ln ( x )}{x ^{3}} d x

\int_{- 1}^{1} e^{x} - x^{2} + 1 d x

\int \frac{x ^{2}}{x ^{2} + x ^{4}} d x

x^{2} y d x + x^{3} d y = 0

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)