What is the solution for 25y^{''}-100y^'+109y=0,y(0)=9,y^'(0)=8 ?

The solution for 25y^{''}-100y^'+109y=0,y(0)=9,y^'(0)=8 is y=e^{2t}(9cos((3t)/5)-50/3 sin((3t)/5))

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25y^{''}-100y^'+109y=0,y(0)=9,y^'(0)=8

25 y^{^{''}} - 100 y^{^{'}} + 109 y = 0, y (0) = 9, y^{^{'}} (0) = 8

y = e^{2 t} (9 cos (\frac{3 t}{5}) - \frac{50}{3} sin (\frac{3 t}{5}))

Solution steps

25 y^{''} (t) - 100 y^{'} (t) + 109 y = 0

Solve linear ODE:

y = e^{2 t} (9 cos (\frac{3 t}{5}) - \frac{50}{3} sin (\frac{3 t}{5}))

y = e^{2 t} (9 cos (\frac{3 t}{5}) - \frac{50}{3} sin (\frac{3 t}{5}))

256 y^{^{''''}} = 81 y

3 y^{^{''}} + 15 y = 0

y^{^{''}} + (2 a - 5) y^{^{'}} + a (a - 5) y = 0

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

\frac{1}{4} x^{^{''}} + 10 x^{^{'}} + 1000 x = 0

What is the solution for 25y^{''}-100y^'+109y=0,y(0)=9,y^'(0)=8 ?
The solution for 25y^{''}-100y^'+109y=0,y(0)=9,y^'(0)=8 is y=e^{2t}(9cos((3t)/5)-50/3 sin((3t)/5))