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

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

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

y^{^{''}} - 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

y^{''} (x) - 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^{^{''}} + 9 y = 5 csc^{2} (3 x)

y^{^{''}} - y - 12 y = e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x}

6 y^{^{''}} - y^{^{'}} - 2 y = 3 sin (10) x

d, d (x) = \frac{3 x - 2}{x + 1}

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