What is the solution for y^{''}+8y^'+32y=32,y(0)=2,y^'(0)=-20 ?

The solution for y^{''}+8y^'+32y=32,y(0)=2,y^'(0)=-20 is y=e^{-4t}(cos(4t)-4sin(4t))+1

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y^{''}+8y^'+32y=32,y(0)=2,y^'(0)=-20

y^{^{''}} + 8 y^{^{'}} + 32 y = 32, y (0) = 2, y^{^{'}} (0) = - 20

y = e^{- 4 t} (cos (4 t) - 4 sin (4 t)) + 1

Solution steps

y^{''} (t) + 8 y^{'} (t) + 32 y = 32

Solve linear ODE:

y = e^{- 4 t} (cos (4 t) - 4 sin (4 t)) + 1

y = e^{- 4 t} (cos (4 t) - 4 sin (4 t)) + 1

\int cos (x) cos (2 x) cos (3 x) d x

\int_{0}^{\frac{3}{2} 3} \frac{x ^{3}}{9 - x ^{2}} d x

y = \frac{e ^{x} - e ^{- x}}{2}

\int x^{3} + 3 x - 2 d x

\frac{d}{d t} (\frac{1}{3} t^{3} + \frac{1}{2} t^{2})

What is the solution for y^{''}+8y^'+32y=32,y(0)=2,y^'(0)=-20 ?
The solution for y^{''}+8y^'+32y=32,y(0)=2,y^'(0)=-20 is y=e^{-4t}(cos(4t)-4sin(4t))+1