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

The solution for y^{''}-y^'=cos(2x) is y=c_{1}e^x+c_{2}-1/10 sin(2x)-1/5 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

y = c_{1} e^{x} + c_{2} - \frac{1}{10} sin (2 x) - \frac{1}{5} cos (2 x)

x^{^{''}} + 16 x = sin (4 t)

y^{^{''}} + 3 y^{^{'}} - 4 y = 2 (5 x + 1) e^{x}

y^{^{''}} + 9 y = 8 cos (3 x) + 12 sin (3 x)

2 y^{^{''}} + 27 y = e^{t} - e^{- t}, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - y^{^{'}} + 2 y = 6 e^{\frac{1}{2}}

What is the solution for y^{''}-y^'=cos(2x) ?
The solution for y^{''}-y^'=cos(2x) is y=c_{1}e^x+c_{2}-1/10 sin(2x)-1/5 cos(2x)