What is the solution for 41y-8y^'+y^{''}=0,y(0)=-3,y^'(0)=-32 ?

The solution for 41y-8y^'+y^{''}=0,y(0)=-3,y^'(0)=-32 is y=e^{4t}(-3cos(5t)-4sin(5t))

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41y-8y^'+y^{''}=0,y(0)=-3,y^'(0)=-32

41 y - 8 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = - 32

y = e^{4 t} (- 3 cos (5 t) - 4 sin (5 t))

Solution steps

41 y - 8 y^{'} (t) + y^{''} (t) = 0

Solve linear ODE:

y = e^{4 t} (- 3 cos (5 t) - 4 sin (5 t))

y = e^{4 t} (- 3 cos (5 t) - 4 sin (5 t))

What is the solution for 41y-8y^'+y^{''}=0,y(0)=-3,y^'(0)=-32 ?
The solution for 41y-8y^'+y^{''}=0,y(0)=-3,y^'(0)=-32 is y=e^{4t}(-3cos(5t)-4sin(5t))