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

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

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y^{'''}-49y^'=0

y^{^{'''}} - 49 y^{^{'}} = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} = - y

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = e^{- \frac{π}{6}}

\frac{d ^{3} y}{d t ^{3}} - \frac{d ^{2} y}{d t ^{2}} = 0

y^{^{''}} + 4 y = 0, y (0) = 0, y^{^{'}} (0) = \frac{18}{5}

64 y^{^{''}} + 16 y^{^{'}} + 50 y = 0

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