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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - 40 y^{^{'''}} + 400 y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} + 11 \frac{d y}{d x} + 24 y = 0

x^{^{''}} + 2 x^{^{'}} + 8 x = 0

y^{^{''}} - 15 y^{^{'}} + 50 y = 0

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

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