What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=-7 ?

The solution for y^{''}+9y=0,y(0)=5,y^'(0)=-7 is y=5cos(3t)-7/3 sin(3t)

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y^{''}+9y=0,y(0)=5,y^'(0)=-7

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = - 7

y = 5 cos (3 t) - \frac{7}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

y = 5 cos (3 t) - \frac{7}{3} sin (3 t)

y = 5 cos (3 t) - \frac{7}{3} sin (3 t)

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

2 y = 3 y^{^{''}}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

x^{^{'''}} + 5 x^{^{''}} - 25 x^{^{'}} - 125 x = 0

What is the solution for y^{''}+9y=0,y(0)=5,y^'(0)=-7 ?
The solution for y^{''}+9y=0,y(0)=5,y^'(0)=-7 is y=5cos(3t)-7/3 sin(3t)