What is the solution for y^{''}+4y^'=17cos(t),y(0)=-1,y^'(0)=0 ?

The solution for y^{''}+4y^'=17cos(t),y(0)=-1,y^'(0)=0 is y=-1+e^{-4t}+4sin(t)-cos(t)

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y^{''}+4y^'=17cos(t),y(0)=-1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} = 17 cos (t), y (0) = - 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) = 17 cos (t)

Solve linear ODE:

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

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

\frac{d}{d x} (- \frac{16}{( 1 + 4 x ) ^{2}})

\int_{1}^{2} 21 t^{2} d t

\frac{d}{d x} (x - 8)

4^{\frac{x}{l n ( 2 )}}, at x = ln (16)

\int t (t^{2} + 1) d t

What is the solution for y^{''}+4y^'=17cos(t),y(0)=-1,y^'(0)=0 ?
The solution for y^{''}+4y^'=17cos(t),y(0)=-1,y^'(0)=0 is y=-1+e^{-4t}+4sin(t)-cos(t)