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

The solution for y^{''}+2y^'+y=cos(3x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+3/50 sin(3x)-2/25 cos(3x)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} + \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

What is the solution for y^{''}+2y^'+y=cos(3x) ?
The solution for y^{''}+2y^'+y=cos(3x) is y=c_{1}e^{-x}+c_{2}xe^{-x}+3/50 sin(3x)-2/25 cos(3x)