What is the solution for 4y^{''}+20y^'+169y=0,y(0)=3,y^'(0)=1 ?

The solution for 4y^{''}+20y^'+169y=0,y(0)=3,y^'(0)=1 is y=e^{-(5t)/2}(3cos(6t)+17/12 sin(6t))

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4y^{''}+20y^'+169y=0,y(0)=3,y^'(0)=1

4 y^{^{''}} + 20 y^{^{'}} + 169 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y = e^{- \frac{5 t}{2}} (3 cos (6 t) + \frac{17}{12} sin (6 t))

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 169 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (3 cos (6 t) + \frac{17}{12} sin (6 t))

y = e^{- \frac{5 t}{2}} (3 cos (6 t) + \frac{17}{12} sin (6 t))

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

y^{^{'''}} + 4 y^{^{''}} - 12 y^{^{'}} = 0

y^{^{''}} - 6 y^{^{'}} - 9 y = 0

16 y^{^{''}} - 72 y^{^{'}} + 81 y = 0

y^{^{''''}} - 2 y^{^{''}} - 8 y = 0

What is the solution for 4y^{''}+20y^'+169y=0,y(0)=3,y^'(0)=1 ?
The solution for 4y^{''}+20y^'+169y=0,y(0)=3,y^'(0)=1 is y=e^{-(5t)/2}(3cos(6t)+17/12 sin(6t))