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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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