What is the solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 ?

The solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(5t)+1/5 sin(5t))

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y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 4 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

\int_{- 1}^{1} \frac{1}{x ^{\frac{2}{3}}} d x

\frac{d}{d x} ((\frac{x + 1}{x})^{3})

\frac{d}{d x} (\frac{x ^{2}}{x ^{2} - 16})

\int x^{9} ln (x) d x

f (x) = (13 x - 1)^{- \frac{1}{3}}, at x = 5

What is the solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 ?
The solution for y^{''}+8y^'+41y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(5t)+1/5 sin(5t))