What is the solution for (d-5d^2+4d-20)x=0 ?

The solution for (d-5d^2+4d-20)x=0 is x=e^{t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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(d-5d^2+4d-20)x=0

(d - 5 d^{2} + 4 d - 20) x = 0

x = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

(d - 5 d^{2} + 4 d - 20) x = 0

Solve linear ODE:

x = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

x = e^{\frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

25 x^{^{''}} + 50 x^{^{'}} + 169 x = 0

45 y^{^{''}} + 180 y = 0

\frac{1}{8} x^{^{''}} + 25 x = 0

10 y^{^{''}} + 90 y^{^{'}} + 140 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 3 y = 0

What is the solution for (d-5d^2+4d-20)x=0 ?
The solution for (d-5d^2+4d-20)x=0 is x=e^{t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))