What is the solution for 9y^{''}-6y^'+y=50sin(x) ?

The solution for 9y^{''}-6y^'+y=50sin(x) is y=c_{1}e^{x/3}+c_{2}e^{x/3}x-4sin(x)+3cos(x)

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9y^{''}-6y^'+y=50sin(x)

9 y^{^{''}} - 6 y^{^{'}} + y = 50 sin (x)

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{\frac{x}{3}} x - 4 sin (x) + 3 cos (x)

Solution steps

9 y^{''} (x) - 6 y^{'} (x) + y = 50 sin (x)

Solve linear ODE:

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{\frac{x}{3}} x - 4 sin (x) + 3 cos (x)

y = c_{1} e^{\frac{x}{3}} + c_{2} e^{\frac{x}{3}} x - 4 sin (x) + 3 cos (x)

4 y^{^{''}} + 4 y^{^{'}} + y = x e^{\cdot \frac{x}{2}} sin (x)

y^{^{''}} - y^{^{'}} = 2 x^{2} - 3

y^{^{''}} + 2 y^{^{'}} + y = e^{- x l n (x)}

y^{^{''}} - 8 y^{^{'}} + 80 y = - 3 sin (3 t)

y^{^{''}} - 8 y^{^{'}} + 16 = (1 - x) e^{4 x}

What is the solution for 9y^{''}-6y^'+y=50sin(x) ?
The solution for 9y^{''}-6y^'+y=50sin(x) is y=c_{1}e^{x/3}+c_{2}e^{x/3}x-4sin(x)+3cos(x)