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

The solution for y^{''}+2y^'+10y=0,y(pi/2)=0 is y=c_{1}e^{-t}cos(3t)

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

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

y = c_{1} e^{- t} cos (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} cos (3 t)

y = c_{1} e^{- t} cos (3 t)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 3, y^{^{'}} (- 2) = 5

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

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

y^{^{''}} - 4 y^{^{'}} - a^{2} y = 0

What is the solution for y^{''}+2y^'+10y=0,y(pi/2)=0 ?
The solution for y^{''}+2y^'+10y=0,y(pi/2)=0 is y=c_{1}e^{-t}cos(3t)