What is the solution for 25x^{''}-20x^'+4x=0 ?

The solution for 25x^{''}-20x^'+4x=0 is x=c_{1}e^{(2t)/5}+c_{2}e^{(2t)/5}t

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25x^{''}-20x^'+4x=0

25 x^{^{''}} - 20 x^{^{'}} + 4 x = 0

x = c_{1} e^{\frac{2 t}{5}} + c_{2} e^{\frac{2 t}{5}} t

Solution steps

25 x^{''} (t) - 20 x^{'} (t) + 4 x = 0

Solve linear ODE:

x = c_{1} e^{\frac{2 t}{5}} + c_{2} e^{\frac{2 t}{5}} t

x = c_{1} e^{\frac{2 t}{5}} + c_{2} e^{\frac{2 t}{5}} t

\frac{d ^{2} y}{d x ^{2}} = - k y

2 y^{^{''}} + 5 y^{^{'}} + 2 y = 0

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (0) = 2, y^{^{'}} (0) = 3

(d^{5} + 8 d^{3} + 16 d) y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 1, y (1) = 0

What is the solution for 25x^{''}-20x^'+4x=0 ?
The solution for 25x^{''}-20x^'+4x=0 is x=c_{1}e^{(2t)/5}+c_{2}e^{(2t)/5}t