What is the solution for x^{''}-c^2x=ccos(ct+a) ?

The solution for x^{''}-c^2x=ccos(ct+a) is x=c_{1}e^{ct}+c_{2}e^{-ct}-(cos(ct+a))/(2c)

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x^{''}-c^2x=ccos(ct+a)

x^{^{''}} - c^{2} x = c cos (c t + a)

x = c_{1} e^{c t} + c_{2} e^{- c t} - \frac{cos ( c t + a )}{2 c}

Solution steps

x^{''} (t) - c^{2} x = c cos (c t + a)

Solve linear ODE:

x = c_{1} e^{c t} + c_{2} e^{- c t} - \frac{cos ( c t + a )}{2 c}

x = c_{1} e^{c t} + c_{2} e^{- c t} - \frac{cos ( c t + a )}{2 c}

x^{^{''}} - 3 x^{^{'}} + 2 x = \frac{( e ^{2 t} )}{e ^{2 t} + 1}

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 3 e^{3 x}

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

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

y^{^{'''}} + 6 y^{^{''}} = 3 - cos (x)

What is the solution for x^{''}-c^2x=ccos(ct+a) ?
The solution for x^{''}-c^2x=ccos(ct+a) is x=c_{1}e^{ct}+c_{2}e^{-ct}-(cos(ct+a))/(2c)