What is the solution for y^{''}+9y=0,y(0)=10,y^'(0)=0 ?

The solution for y^{''}+9y=0,y(0)=10,y^'(0)=0 is y=10cos(3t)

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y^{''}+9y=0,y(0)=10,y^'(0)=0

y^{^{''}} + 9 y = 0, y (0) = 10, y^{^{'}} (0) = 0

y = 10 cos (3 t)

Solution steps

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

Solve linear ODE:

y = 10 cos (3 t)

y = 10 cos (3 t)

4 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 3 y = 0

y^{^{''''}} + 2 y^{^{'''}} + 37 y^{^{''}} + 72 y^{^{'}} + 36 y = 0

\frac{d ^{2} y}{d x ^{2}} + y = 0, y = 1, y^{^{'}} = 0

y^{^{'''}} + 3.2 y^{^{''}} + 4.81 y^{^{'}} = 0

3 y^{^{''}} + 27 y^{^{'}} + 42 y = 0

What is the solution for y^{''}+9y=0,y(0)=10,y^'(0)=0 ?
The solution for y^{''}+9y=0,y(0)=10,y^'(0)=0 is y=10cos(3t)