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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

sec^{2} (x)

2 y^{^{''}} + 17 y^{^{'}} - 9 y = 0

\frac{d}{d y} (x y + y z + x z)

y^{^{'}} = 3 e^{x - y}

f (x) = tan (4 x)

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