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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = e^{2 x}, y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} + y = cos (t)

y^{^{''}} + y = 7 x sin (x)

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

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

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