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

The solution for y^{''}-y^'=3cos(2x) is y=c_{1}e^x+c_{2}-3/10 sin(2x)-3/5 cos(2x)

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

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

y = c_{1} e^{x} + c_{2} - \frac{3}{10} sin (2 x) - \frac{3}{5} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} - \frac{3}{10} sin (2 x) - \frac{3}{5} cos (2 x)

y = c_{1} e^{x} + c_{2} - \frac{3}{10} sin (2 x) - \frac{3}{5} cos (2 x)

What is the solution for y^{''}-y^'=3cos(2x) ?
The solution for y^{''}-y^'=3cos(2x) is y=c_{1}e^x+c_{2}-3/10 sin(2x)-3/5 cos(2x)