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

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

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

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

y = 9 cos (3 t) - 3 sin (3 t)

Solution steps

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

Solve linear ODE:

y = 9 cos (3 t) - 3 sin (3 t)

y = 9 cos (3 t) - 3 sin (3 t)

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

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = 1, y (0) = 7

2 y^{^{''}} + 9 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

\frac{d ^{2} x}{d t ^{2}} + 2 a \frac{d x}{d t} + b^{2} x = 0, a < b

x^{^{''}} + 4 x^{^{'}} + 5 x = 0, x (0) = 0, x^{^{'}} (0) = 4

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