What is the solution for 45y^{''}+180y=0 ?

The solution for 45y^{''}+180y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)

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45y^{''}+180y=0

45 y^{^{''}} + 180 y = 0

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

Solution steps

45 y^{''} (t) + 180 y = 0

Solve linear ODE:

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

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

\frac{1}{8} x^{^{''}} + 25 x = 0

10 y^{^{''}} + 90 y^{^{'}} + 140 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 3 y = 0

\frac{d ^{2} x}{d t ^{2}} + 6 \frac{d x}{d t} + 9 x = 0

y^{^{'''}} + y^{^{'}} = 0, y^{^{'}} (0) = 1, y^{^{''}} (0) = 2

What is the solution for 45y^{''}+180y=0 ?
The solution for 45y^{''}+180y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)