What is the solution for 3y^{''}-24y^'+36y=6sin(2x) ?

The solution for 3y^{''}-24y^'+36y=6sin(2x) is y=c_{1}e^{6x}+c_{2}e^{2x}+1/20 sin(2x)+1/10 cos(2x)

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3y^{''}-24y^'+36y=6sin(2x)

3 y^{^{''}} - 24 y^{^{'}} + 36 y = 6 sin (2 x)

y = c_{1} e^{6 x} + c_{2} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

Solution steps

3 y^{''} (x) - 24 y^{'} (x) + 36 y = 6 sin (2 x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

y = c_{1} e^{6 x} + c_{2} e^{2 x} + \frac{1}{20} sin (2 x) + \frac{1}{10} cos (2 x)

\frac{d}{d x} (\frac{1}{1 + x ^{8}})

\int 6 e^{- 0.5 x} d x

\frac{\partial}{\partial x} (ln (5 x^{2} - 2 y^{2}))

\int_{1}^{3} \frac{2}{4 + ( x - 1 ) ^{2}} d x

y^{^{'}} = \frac{2 y}{x}

What is the solution for 3y^{''}-24y^'+36y=6sin(2x) ?
The solution for 3y^{''}-24y^'+36y=6sin(2x) is y=c_{1}e^{6x}+c_{2}e^{2x}+1/20 sin(2x)+1/10 cos(2x)