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

The solution for y^{''}-2y^'=sin(4x) is y=c_{1}e^{2x}+c_{2}-1/20 sin(4x)+1/40 cos(4x)

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

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

y = c_{1} e^{2 x} + c_{2} - \frac{1}{20} sin (4 x) + \frac{1}{40} cos (4 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} - \frac{1}{20} sin (4 x) + \frac{1}{40} cos (4 x)

y = c_{1} e^{2 x} + c_{2} - \frac{1}{20} sin (4 x) + \frac{1}{40} cos (4 x)

y^{^{''}} + 4 y = 4 cos (x) + 3 sin (x) - 8

16 y^{^{''}} - 40 y + 25 = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 12

y^{^{''}} + 4 y = 2 tan (2 x)

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (x)

What is the solution for y^{''}-2y^'=sin(4x) ?
The solution for y^{''}-2y^'=sin(4x) is y=c_{1}e^{2x}+c_{2}-1/20 sin(4x)+1/40 cos(4x)