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

The solution for y^{''}-4y^'+3y=cos(4t) is y=c_{1}e^{3t}+c_{2}e^t-16/425 sin(4t)-13/425 cos(4t)

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

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

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{16}{425} sin (4 t) - \frac{13}{425} cos (4 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{16}{425} sin (4 t) - \frac{13}{425} cos (4 t)

y = c_{1} e^{3 t} + c_{2} e^{t} - \frac{16}{425} sin (4 t) - \frac{13}{425} cos (4 t)

y = \frac{x}{x + 3}

\frac{d}{d x} (cos (\frac{x}{4}))

y^{^{''}} - 4 y^{^{'}} + 4 y = (3 x + 8) e^{2 x} + 8 x^{2}

\int y ln (y) - \frac{1}{9 - y ^{2}} d y

\frac{\partial}{\partial x} (2 x - 10)

What is the solution for y^{''}-4y^'+3y=cos(4t) ?
The solution for y^{''}-4y^'+3y=cos(4t) is y=c_{1}e^{3t}+c_{2}e^t-16/425 sin(4t)-13/425 cos(4t)