What is the solution for 4y^{''}+3y=0 ?

The solution for 4y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2)

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4y^{''}+3y=0

4 y^{^{''}} + 3 y = 0

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

Solution steps

4 y^{''} (t) + 3 y = 0

Solve linear ODE:

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

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

\int 8 x - 3 d x

\int_{1}^{2} \frac{x ^{2} - x - 5}{x + 2} d x

y = \frac{4}{x}, y = \frac{4}{x ^{2}}, x = 3

\int x ln (13 x) d x

\frac{9}{22 ( 3 s - 2 )}

What is the solution for 4y^{''}+3y=0 ?
The solution for 4y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2)