What is the solution for y^{''}-6y^'+8y=3sin(2x) ?

The solution for y^{''}-6y^'+8y=3sin(2x) is y=c_{1}e^{4x}+c_{2}e^{2x}+3/40 sin(2x)+9/40 cos(2x)

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y^{''}-6y^'+8y=3sin(2x)

y^{^{''}} - 6 y^{^{'}} + 8 y = 3 sin (2 x)

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3}{40} sin (2 x) + \frac{9}{40} cos (2 x)

Solution steps

y^{''} (x) - 6 y^{'} (x) + 8 y = 3 sin (2 x)

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3}{40} sin (2 x) + \frac{9}{40} cos (2 x)

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{3}{40} sin (2 x) + \frac{9}{40} cos (2 x)

What is the solution for y^{''}-6y^'+8y=3sin(2x) ?
The solution for y^{''}-6y^'+8y=3sin(2x) is y=c_{1}e^{4x}+c_{2}e^{2x}+3/40 sin(2x)+9/40 cos(2x)