What is the solution for 25y^{''}-40y^'+16y=0 ?

The solution for 25y^{''}-40y^'+16y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{(4t)/5}t

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25y^{''}-40y^'+16y=0

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{\frac{4 t}{5}} t

Solution steps

25 y^{''} (t) - 40 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{\frac{4 t}{5}} t

y = c_{1} e^{\frac{4 t}{5}} + c_{2} e^{\frac{4 t}{5}} t

6 y^{^{''}} + y^{^{'}} - 5 y = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 13 y = 0

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

6 y^{^{''}} - 5 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = 1

(D^{2} + 9 D + 18) y = 0

What is the solution for 25y^{''}-40y^'+16y=0 ?
The solution for 25y^{''}-40y^'+16y=0 is y=c_{1}e^{(4t)/5}+c_{2}e^{(4t)/5}t