What is the solution for 4y^{''}-28y^'+49y=0 ?

The solution for 4y^{''}-28y^'+49y=0 is y=c_{1}e^{(7t)/2}+c_{2}e^{(7t)/2}t

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4y^{''}-28y^'+49y=0

4 y^{^{''}} - 28 y^{^{'}} + 49 y = 0

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

Solution steps

4 y^{''} (t) - 28 y^{'} (t) + 49 y = 0

Solve linear ODE:

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

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

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

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

\frac{d ^{2} f}{d x ^{2}} = - a^{2} f (x)

\frac{d ^{2} y}{d x ^{2}} - 36 y = 0

2 y^{^{''}} + y^{^{'}} + 3 y = 0

What is the solution for 4y^{''}-28y^'+49y=0 ?
The solution for 4y^{''}-28y^'+49y=0 is y=c_{1}e^{(7t)/2}+c_{2}e^{(7t)/2}t