What is the solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=8 ?

The solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=8 is y=-8e^{(3pi)/2-3t}cos(t)

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y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=8

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 8

y = - 8 e^{\frac{3 π}{2} - 3 t} cos (t)

Solution steps

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

Solve linear ODE:

y = - 8 e^{\frac{3 π}{2} - 3 t} cos (t)

y = - 8 e^{\frac{3 π}{2} - 3 t} cos (t)

(d^{2} - 2) y = 0

\frac{d ^{6} y}{d x ^{6}} - 2 \frac{d ^{3} y}{d x ^{3}} + y = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 4, y^{^{'}} (0) = - 10

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

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = - 4, y^{^{'}} (0) = 3

What is the solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=8 ?
The solution for y^{''}+6y^'+10y=0,y(pi/2)=0,y^'(pi/2)=8 is y=-8e^{(3pi)/2-3t}cos(t)