What is the solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=2 ?

The solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-2e^{(3pi)/2-3t}cos(t)

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y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=2

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 2

y = - 2 e^{\frac{3 π}{2} - 3 t} cos (t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = - 2 e^{\frac{3 π}{2} - 3 t} cos (t)

y = - 2 e^{\frac{3 π}{2} - 3 t} cos (t)

y^{^{''}} + \frac{y}{4} = 0

y^{^{''}} - y^{^{'}} + 144 y = 0

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

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} - 22 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (10) + 25 y = 0

What is the solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=2 ?
The solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-2e^{(3pi)/2-3t}cos(t)