What is the solution for y^{''}+2y^'+17y=20e^t,y(0)=2,y^{''}(0)=-4 ?

The solution for y^{''}+2y^'+17y=20e^t,y(0)=2,y^{''}(0)=-4 is y=e^{-t}(cos(4t)-5/4 sin(4t))+e^t

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y^{''}+2y^'+17y=20e^t,y(0)=2,y^{''}(0)=-4

y^{^{''}} + 2 y^{^{'}} + 17 y = 20 e^{t}, y (0) = 2, y^{^{''}} (0) = - 4

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 y = 20 e^{t}

Solve linear ODE:

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

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

y^{^{''}} + 7 y^{^{'}} + 12 y = 2 e^{- t}

y^{^{''}} + 16 y = sin (8 x)

2 x^{^{''}} + 3 x^{^{'}} + 5 x = 2 t

y^{^{''}} - 6 y^{^{'}} + 8 y = 8 e^{4 x}

2 y^{^{''}} + 18 y = 3 tan (3 x)

What is the solution for y^{''}+2y^'+17y=20e^t,y(0)=2,y^{''}(0)=-4 ?
The solution for y^{''}+2y^'+17y=20e^t,y(0)=2,y^{''}(0)=-4 is y=e^{-t}(cos(4t)-5/4 sin(4t))+e^t