What is the solution for 3y^{''}-30y^'+72y=6sin(4x) ?

The solution for 3y^{''}-30y^'+72y=6sin(4x) is y=c_{1}e^{6x}+c_{2}e^{4x}+1/104 sin(4x)+5/104 cos(4x)

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3y^{''}-30y^'+72y=6sin(4x)

3 y^{^{''}} - 30 y^{^{'}} + 72 y = 6 sin (4 x)

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{104} sin (4 x) + \frac{5}{104} cos (4 x)

Solution steps

3 y^{''} (x) - 30 y^{'} (x) + 72 y = 6 sin (4 x)

Solve linear ODE:

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{104} sin (4 x) + \frac{5}{104} cos (4 x)

y = c_{1} e^{6 x} + c_{2} e^{4 x} + \frac{1}{104} sin (4 x) + \frac{5}{104} cos (4 x)

\int \frac{3 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{d}{d x} (sin (x^{3}) + 3 x y)

f (x) = \frac{x}{x + 3}, at x = 1

f (x) = e^{6 x} cos (9 x)

\frac{d}{d x} (ln (2 + x^{2}))

What is the solution for 3y^{''}-30y^'+72y=6sin(4x) ?
The solution for 3y^{''}-30y^'+72y=6sin(4x) is y=c_{1}e^{6x}+c_{2}e^{4x}+1/104 sin(4x)+5/104 cos(4x)