What is the solution for y^{'''}+7y^'=0 ?

The solution for y^{'''}+7y^'=0 is y=c_{1}+c_{2}cos(sqrt(7)t)+c_{3}sin(sqrt(7)t)

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y^{'''}+7y^'=0

y^{^{'''}} + 7 y^{^{'}} = 0

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

Solution steps

y^{'''} (t) + 7 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

y = c_{1} + c_{2} cos (7 t) + c_{3} sin (7 t)

x^{^{''}} = b x

y^{^{''}} + \frac{16}{3} y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{4}{3}

2 y^{^{''}} + 5 y^{^{'}} = 0

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{1} (0) = - x e^{x}

What is the solution for y^{'''}+7y^'=0 ?
The solution for y^{'''}+7y^'=0 is y=c_{1}+c_{2}cos(sqrt(7)t)+c_{3}sin(sqrt(7)t)