What is the solution for y^{''}-32y=16cos(x) ?

The solution for y^{''}-32y=16cos(x) is y=c_{1}e^{4sqrt(2)x}+c_{2}e^{-4sqrt(2)x}-16/33 cos(x)

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y^{''}-32y=16cos(x)

y^{^{''}} - 32 y = 16 cos (x)

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{16}{33} cos (x)

Solution steps

y^{''} (x) - 32 y = 16 cos (x)

Solve linear ODE:

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{16}{33} cos (x)

y = c_{1} e^{42 x} + c_{2} e^{- 42 x} - \frac{16}{33} cos (x)

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 91

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

x^{^{''}} + 16 x = cos (4 t), x (0) = 0, x^{^{'}} (0) = 1

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{4} e^{t}

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

What is the solution for y^{''}-32y=16cos(x) ?
The solution for y^{''}-32y=16cos(x) is y=c_{1}e^{4sqrt(2)x}+c_{2}e^{-4sqrt(2)x}-16/33 cos(x)