What is the solution for 16y^{''}-56y^'+49y=0 ?

The solution for 16y^{''}-56y^'+49y=0 is y=c_{1}e^{(7t)/4}+c_{2}e^{(7t)/4}t

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16y^{''}-56y^'+49y=0

16 y^{^{''}} - 56 y^{^{'}} + 49 y = 0

y = c_{1} e^{\frac{7 t}{4}} + c_{2} e^{\frac{7 t}{4}} t

Solution steps

16 y^{''} (t) - 56 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{7 t}{4}} + c_{2} e^{\frac{7 t}{4}} t

y = c_{1} e^{\frac{7 t}{4}} + c_{2} e^{\frac{7 t}{4}} t

y^{^{''}} = 4 y, y (0) = 1, y (1) = 0

y^{^{''}} - 4 a^{2} y^{^{'}} + y = 0

y^{^{''}} + 36 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 16

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{2}

What is the solution for 16y^{''}-56y^'+49y=0 ?
The solution for 16y^{''}-56y^'+49y=0 is y=c_{1}e^{(7t)/4}+c_{2}e^{(7t)/4}t