What is the solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=9 ?

The solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=9 is y=3e^{pi/2-t}cos(3t)

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y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=9

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

y = 3 e^{\frac{π}{2} - t} cos (3 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = 3 e^{\frac{π}{2} - t} cos (3 t)

y = 3 e^{\frac{π}{2} - t} cos (3 t)

6 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

x^{^{''}} + 2 x^{^{'}} + 9 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} = 0

y^{^{''}} + 12 y^{^{'}} + 16 y = 0

What is the solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=9 ?
The solution for y^{''}+2y^'+10y=0,y(pi/2)=0,y^'(pi/2)=9 is y=3e^{pi/2-t}cos(3t)