What is the solution for (D^2-4)y=cos^2(x) ?

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

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(D^2-4)y=cos^2(x)

(D^{2} - 4) y = cos^{2} (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{cos ( 2 x ) - 4 cos ^{2} ( x )}{16}

Solution steps

(D^{2} - 4) y = cos^{2} (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{cos ( 2 x ) - 4 cos ^{2} ( x )}{16}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{cos ( 2 x ) - 4 cos ^{2} ( x )}{16}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{2 t} + 10 t^{2}

2 y^{^{''}} - 3 y = x + 1

y^{^{''}} - 5 y^{^{'}} + 6 y = x

\frac{1}{6} y^{^{''}} + 6 y = 3 tan (6 t) - \frac{1}{3} e^{t}

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

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