What is the solution for y^{''}+8y^'+20y=40,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+8y^'+20y=40,y(0)=0,y^'(0)=0 is y=e^{-4t}(-2cos(2t)-4sin(2t))+2

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y^{''}+8y^'+20y=40,y(0)=0,y^'(0)=0

y^{^{''}} + 8 y^{^{'}} + 20 y = 40, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 20 y = 40

Solve linear ODE:

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

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

\int csc (t) d t

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

f (x) = 2 x^{3} - 8 x, (- 2, 0)

\int ((4 x + 3) \cdot (2 x^{2} + 3 x)^{3}) d x

f (x) = x \cdot sin (x)

What is the solution for y^{''}+8y^'+20y=40,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+8y^'+20y=40,y(0)=0,y^'(0)=0 is y=e^{-4t}(-2cos(2t)-4sin(2t))+2