What is the solution for y^{''}-6y^'+9y=cos(x) ?

The solution for y^{''}-6y^'+9y=cos(x) is y=c_{1}e^{3x}+c_{2}xe^{3x}-3/50 sin(x)+2/25 cos(x)

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y^{''}-6y^'+9y=cos(x)

y^{^{''}} - 6 y^{^{'}} + 9 y = cos (x)

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

Solution steps

y^{''} (x) - 6 y^{'} (x) + 9 y = cos (x)

Solve linear ODE:

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

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

y^{^{''}} - 4 y = cosh (2 x)

y^{^{''}} - 3 y^{^{'}} = 8 e^{3 x} + 4 sin (x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 18 t^{2}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{t}, y (0) = 1, y^{^{'}} (0) = - 6

y^{^{''}} + 2 y^{^{'}} + 4 y = 111 e^{2 t} cos (3 t)

What is the solution for y^{''}-6y^'+9y=cos(x) ?
The solution for y^{''}-6y^'+9y=cos(x) is y=c_{1}e^{3x}+c_{2}xe^{3x}-3/50 sin(x)+2/25 cos(x)