What is the solution for y^{''}+9y=0,y(0)=3,y^'(0)=-4 ?

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

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y^{''}+9y=0,y(0)=3,y^'(0)=-4

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

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

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

y^{^{''}} = 5 y, y (0) = - 2, y (2) = 10

y^{^{''}} + 9 y = 0, y^{1} (0) = sin (3 x)

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