What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=1 ?

The solution for y^{''}+9y=0,y(0)=5,y^'(0)=1 is y=5cos(3t)+1/3 sin(3t)

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y^{''}+9y=0,y(0)=5,y^'(0)=1

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 9 y = 0

Solve linear ODE:

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

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

x \to 9 lim (\frac{7 x}{x - 9})

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

\int \frac{2 x - 4}{x ^{2} - x} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = 5 e^{- (5 t)}

\int_{0}^{π} 8 sin (x) d x

What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=1 ?
The solution for y^{''}+9y=0,y(0)=5,y^'(0)=1 is y=5cos(3t)+1/3 sin(3t)