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

The solution for y^{''}-3y^'+2y=cos(2x) is y=c_{1}e^{2x}+c_{2}e^x-3/20 sin(2x)-1/20 cos(2x)

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y^{''}-3y^'+2y=cos(2x)

y^{^{''}} - 3 y^{^{'}} + 2 y = cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = cos (2 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{x} - \frac{3}{20} sin (2 x) - \frac{1}{20} cos (2 x)

y^{^{'''}} - y^{^{'}} = 10 cos (2 x)

y^{^{''}} + 2 y^{^{'}} + 2 y = x^{3} - 1

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

x^{^{''}} + 2 x^{^{'}} + 5 x = 8 cos (2 t) + 2 sin (2 t)

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

What is the solution for y^{''}-3y^'+2y=cos(2x) ?
The solution for y^{''}-3y^'+2y=cos(2x) is y=c_{1}e^{2x}+c_{2}e^x-3/20 sin(2x)-1/20 cos(2x)