What is the solution for y^{''}+4y^'+20y=0,y(pi)=4,y(0)=2 ?

The solution for y^{''}+4y^'+20y=0,y(pi)=4,y(0)=2 is y=e^{-2t}(4e^{2pi}cos(4t)+c_{2}sin(4t))

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y^{''}+4y^'+20y=0,y(pi)=4,y(0)=2

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 4, y (0) = 2

y = e^{- 2 t} (4 e^{2 π} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- 2 t} (4 e^{2 π} cos (4 t) + c_{2} sin (4 t))

y = e^{- 2 t} (4 e^{2 π} cos (4 t) + c_{2} sin (4 t))

y^{^{''}} + 2 y^{^{''}} + y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (9) = 0

7 y^{^{''}} + 34 y^{^{'}} - 5 y = 0

(D^{2} - DD + D^{2}) z = 0

9 y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = - 1

What is the solution for y^{''}+4y^'+20y=0,y(pi)=4,y(0)=2 ?
The solution for y^{''}+4y^'+20y=0,y(pi)=4,y(0)=2 is y=e^{-2t}(4e^{2pi}cos(4t)+c_{2}sin(4t))