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

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

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

y^{^{'}} + y - e^{x} = 0

y = x, y = x^{3}, [- 1, 2]

\int - e^{2 t} d t

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

\frac{\partial}{\partial y} (cos (x^{3} y^{3}))

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