What is the solution for (y^{''})/8+32y=2cos(3t)+4 ?

The solution for (y^{''})/8+32y=2cos(3t)+4 is y=c_{1}cos(16t)+c_{2}sin(16t)+16/247 cos(3t)+1/8

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(y^{''})/8+32y=2cos(3t)+4

\frac{y ^{^{''}}}{8} + 32 y = 2 cos (3 t) + 4

y = c_{1} cos (16 t) + c_{2} sin (16 t) + \frac{16}{247} cos (3 t) + \frac{1}{8}

Solution steps

\frac{y ^{''} ( t )}{8} + 32 y = 2 cos (3 t) + 4

Solve linear ODE:

y = c_{1} cos (16 t) + c_{2} sin (16 t) + \frac{16}{247} cos (3 t) + \frac{1}{8}

y = c_{1} cos (16 t) + c_{2} sin (16 t) + \frac{16}{247} cos (3 t) + \frac{1}{8}

y^{^{''}} - 3 y^{^{'}} + 2 y = x, y (0) = e^{x}

y^{^{''}} + y = 2 cos (x) + x

- 3 y^{^{''}} - 2 y^{^{'}} + y = t^{2} + t - 2 e^{2 t}

x^{^{'''}} - 4 x^{^{''}} + 5 x^{^{'}} - 2 x = 2 t + 3

y^{^{''}} + 4 y = sin (2 x + 3)

What is the solution for (y^{''})/8+32y=2cos(3t)+4 ?
The solution for (y^{''})/8+32y=2cos(3t)+4 is y=c_{1}cos(16t)+c_{2}sin(16t)+16/247 cos(3t)+1/8