What is the solution for y^{''}+3y^'+2y=4cos(x) ?

The solution for y^{''}+3y^'+2y=4cos(x) is y=c_{1}e^{-x}+c_{2}e^{-2x}+6/5 sin(x)+2/5 cos(x)

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y^{''}+3y^'+2y=4cos(x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 cos (x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{6}{5} sin (x) + \frac{2}{5} cos (x)

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 4 cos (x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{6}{5} sin (x) + \frac{2}{5} cos (x)

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{6}{5} sin (x) + \frac{2}{5} cos (x)

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 20 cos (2 t)

y^{^{''}} + 12 y^{^{'}} + 36 y = e^{- 6 t} ln (t)

y^{^{''}} - 4 y^{^{'}} = 2 x^{2} + 5

4 y^{^{''}} + 12 y^{^{'}} + 9 = 0

y^{^{''}} - 16 y = 7 e^{4 x}

What is the solution for y^{''}+3y^'+2y=4cos(x) ?
The solution for y^{''}+3y^'+2y=4cos(x) is y=c_{1}e^{-x}+c_{2}e^{-2x}+6/5 sin(x)+2/5 cos(x)