What is the solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=0 ?

The solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=0 is y=e^{-2t}(2cos(5t)+4/5 sin(5t))

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y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{- 2 t} (2 cos (5 t) + \frac{4}{5} sin (5 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- 2 t} (2 cos (5 t) + \frac{4}{5} sin (5 t))

y = e^{- 2 t} (2 cos (5 t) + \frac{4}{5} sin (5 t))

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = - 1, y^{^{'}} (0) = - 18

\frac{d ^{3} y}{d x ^{3}} - 5 \frac{d y}{d x} = 0

u^{^{''}} + (a^{2} + b) H (t) = 0

y^{^{''}} + 13 y^{^{'}} + 36 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 7 y = 0

What is the solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=0 ?
The solution for y^{''}+4y^'+29y=0,y(0)=2,y^'(0)=0 is y=e^{-2t}(2cos(5t)+4/5 sin(5t))