What is the solution for y^{''}+9y=0,y(0)=4,y^'(0)=1 ?

The solution for y^{''}+9y=0,y(0)=4,y^'(0)=1 is y=4cos(3t)+1/3 sin(3t)

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y^{''}+9y=0,y(0)=4,y^'(0)=1

y^{^{''}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y = 4 cos (3 t) + \frac{1}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

y = 4 cos (3 t) + \frac{1}{3} sin (3 t)

y = 4 cos (3 t) + \frac{1}{3} sin (3 t)

4 y^{^{''}} - 2 y^{^{'}} + y = 0

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = 1

16 \cdot y^{^{''}} + 24 \cdot y^{^{'}} + 73 \cdot y = 0, y (0) = [4]

64 y^{^{''}} - 81 y = 0

y^{^{''}} + 4 y = 0, y (3 π) = - 3, y (0) = - 3

What is the solution for y^{''}+9y=0,y(0)=4,y^'(0)=1 ?
The solution for y^{''}+9y=0,y(0)=4,y^'(0)=1 is y=4cos(3t)+1/3 sin(3t)