What is the solution for 25y^{''}+3y=0 ?

The solution for 25y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/5)+c_{2}sin((sqrt(3)t)/5)

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25y^{''}+3y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{4} 2 π x (8 - x^{\frac{3}{2}}) d x

\int_{2}^{3} \frac{33}{3 - x} d x

cos^{2} (16 t)

\int (x + 1) sin (x^{2} + 2 x + 3) d x

\frac{d x}{d t} = e^{x} \cdot sin (t)

What is the solution for 25y^{''}+3y=0 ?
The solution for 25y^{''}+3y=0 is y=c_{1}cos((sqrt(3)t)/5)+c_{2}sin((sqrt(3)t)/5)