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

The solution for 2y^{''}+3y^'+3=cos(x) is y=c_{1}+c_{2}e^{-(3x)/2}+3/13 sin(x)-2/13 cos(x)-x

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

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

y = c_{1} + c_{2} e^{- \frac{3 x}{2}} + \frac{3}{13} sin (x) - \frac{2}{13} cos (x) - x

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{3 x}{2}} + \frac{3}{13} sin (x) - \frac{2}{13} cos (x) - x

y = c_{1} + c_{2} e^{- \frac{3 x}{2}} + \frac{3}{13} sin (x) - \frac{2}{13} cos (x) - x

y^{^{'''}} - 5 y^{^{''}} + 6 y^{^{'}} = 4 + 6 sin (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = - 4 x e^{- 2 x} + 2 e^{- 3 x} + 3

(d^{2} + 4 d + 5) y = 50 x + 13 e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = e^{- t}

y^{^{''}} + 5 y^{^{'}} - 6 y = 2 e^{- x}

What is the solution for 2y^{''}+3y^'+3=cos(x) ?
The solution for 2y^{''}+3y^'+3=cos(x) is y=c_{1}+c_{2}e^{-(3x)/2}+3/13 sin(x)-2/13 cos(x)-x