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

The solution for y^{'''}+49y^'=0 is y=c_{1}+c_{2}cos(7t)+c_{3}sin(7t)

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

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

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

Solution steps

y^{'''} (t) + 49 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)

t an g e n t e^{x^{5}}

\frac{d}{d x} (8 x^{2} + 7)

\int_{0}^{2} 3^{- x} d x

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{x} cos (3 x)

\int sin (u^{2}) d u

What is the solution for y^{'''}+49y^'=0 ?
The solution for y^{'''}+49y^'=0 is y=c_{1}+c_{2}cos(7t)+c_{3}sin(7t)