What is the solution for y^{''}+8y^'+32y=128,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+8y^'+32y=128,y(0)=0,y^'(0)=0 is y=e^{-4t}(-4cos(4t)-4sin(4t))+4

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y^{''}+8y^'+32y=128,y(0)=0,y^'(0)=0

y^{^{''}} + 8 y^{^{'}} + 32 y = 128, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 32 y = 128

Solve linear ODE:

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

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

\frac{d}{d t} (x y (t) z (t))

y^{^{'}} + tan (x) y = \frac{1}{cos ( x )}

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

x \to 0 - lim (ln ∣ x ∣)

\frac{d y}{d x} + 11 y = 4

What is the solution for y^{''}+8y^'+32y=128,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+8y^'+32y=128,y(0)=0,y^'(0)=0 is y=e^{-4t}(-4cos(4t)-4sin(4t))+4