What is the solution for 9y^{''}-42y^'+49y=0 ?

The solution for 9y^{''}-42y^'+49y=0 is y=c_{1}e^{(7t)/3}+c_{2}e^{(7t)/3}t

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9y^{''}-42y^'+49y=0

9 y^{^{''}} - 42 y^{^{'}} + 49 y = 0

y = c_{1} e^{\frac{7 t}{3}} + c_{2} e^{\frac{7 t}{3}} t

Solution steps

9 y^{''} (t) - 42 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = c_{1} e^{\frac{7 t}{3}} + c_{2} e^{\frac{7 t}{3}} t

y = c_{1} e^{\frac{7 t}{3}} + c_{2} e^{\frac{7 t}{3}} t

x^{^{''}} + 2 a x^{^{'}} + b x = 0

y^{^{''''}} + 4 y^{^{'''}} + 4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

x^{^{''}} - 6 x^{^{''}} + x = 0

y^{^{'''}} + 9 y^{^{'}} = 0, y (0) = 5, y^{^{'}} (0) = 4, y^{^{''}} (0) = 3

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

What is the solution for 9y^{''}-42y^'+49y=0 ?
The solution for 9y^{''}-42y^'+49y=0 is y=c_{1}e^{(7t)/3}+c_{2}e^{(7t)/3}t