What is the solution for y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1 ?

The solution for y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1 is y=e^{-2t}(-2cos(t)-sin(t))-2e^t

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y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 5 y = - 20 e^{t}, y (0) = - 4, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = - 20 e^{t}

Solve linear ODE:

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

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

\int_{1}^{4} 6 e^{x} d x

f (x, y) = (x)^{2} + (y)^{4}

x \to 3 + lim (\frac{1}{∣ x - 3 ∣})

y \to 0 lim (\frac{y ^{2}}{y ^{4}})

f (x) = e^{x} - x^{7}

What is the solution for y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1 ?
The solution for y^{''}+4y^'+5y=-20e^t,y(0)=-4,y^'(0)=1 is y=e^{-2t}(-2cos(t)-sin(t))-2e^t