What is the solution for y^{''}+5y^'+29/4 y=0,y(0)=1,y^'(0)=-7/2 ?

The solution for y^{''}+5y^'+29/4 y=0,y(0)=1,y^'(0)=-7/2 is y=e^{-(5t)/2}(cos(t)-sin(t))

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y^{''}+5y^'+29/4 y=0,y(0)=1,y^'(0)=-7/2

y^{^{''}} + 5 y^{^{'}} + \frac{29}{4} y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{7}{2}

y = e^{- \frac{5 t}{2}} (cos (t) - sin (t))

Solution steps

y^{''} (t) + 5 y^{'} (t) + \frac{29}{4} y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (cos (t) - sin (t))

y = e^{- \frac{5 t}{2}} (cos (t) - sin (t))

y^{^{''''}} - 2 y^{^{''}} = 0

- 10 y^{^{''}} + y = 0

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 2

4 y^{^{''}} - 4 y^{^{'}} + 7 y = 0

(D^{2} + 3 D + 2) y = 0

What is the solution for y^{''}+5y^'+29/4 y=0,y(0)=1,y^'(0)=-7/2 ?
The solution for y^{''}+5y^'+29/4 y=0,y(0)=1,y^'(0)=-7/2 is y=e^{-(5t)/2}(cos(t)-sin(t))