What is the solution for y^{''}+4y^'+20y=0,y(pi)=8,y(0)=1 ?

The solution for y^{''}+4y^'+20y=0,y(pi)=8,y(0)=1 is y=e^{-2t}(8e^{2pi}cos(4t)+c_{2}sin(4t))

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y^{''}+4y^'+20y=0,y(pi)=8,y(0)=1

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 8, y (0) = 1

y = e^{- 2 t} (8 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} (8 e^{2 π} cos (4 t) + c_{2} sin (4 t))

y = e^{- 2 t} (8 e^{2 π} cos (4 t) + c_{2} sin (4 t))

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

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

u^{''} - u^{'} - 2 H (t) = 0

y^{^{''}} + 0.25 y^{^{'}} + y = 0

2 y^{^{'''}} + 5 y^{^{''}} - 28 y^{^{'}} - 15 y = 0

What is the solution for y^{''}+4y^'+20y=0,y(pi)=8,y(0)=1 ?
The solution for y^{''}+4y^'+20y=0,y(pi)=8,y(0)=1 is y=e^{-2t}(8e^{2pi}cos(4t)+c_{2}sin(4t))