What is the solution for y^{''}+4y=6sin(3t) ?

The solution for y^{''}+4y=6sin(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)-6/5 sin(3t)

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y^{''}+4y=6sin(3t)

y^{^{''}} + 4 y = 6 sin (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{6}{5} sin (3 t)

Solution steps

y^{''} (t) + 4 y = 6 sin (3 t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{6}{5} sin (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{6}{5} sin (3 t)

\int 6 sin^{4} (x) cos (x) d x

x^{3} d x + y^{3} d y = 0

\int sin^{5} (t) cos^{4} (t) d t

f^{^{'}} (x) = \frac{1 - x}{x ^{2}}

f (x) = \frac{2}{x + 4}, at x = - 3

What is the solution for y^{''}+4y=6sin(3t) ?
The solution for y^{''}+4y=6sin(3t) is y=c_{1}cos(2t)+c_{2}sin(2t)-6/5 sin(3t)