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

The solution for y^{'''}+4y^'=0 is y=c_{1}+c_{2}cos(2t)+c_{3}sin(2t)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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\frac{d}{d x} (9 e^{4 x})

\frac{\partial}{\partial x} (\frac{2 x ^{2} + 7 x ^{2} y}{3 x y})

x \to 6 lim (\frac{1}{x - 6})

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What is the solution for y^{'''}+4y^'=0 ?
The solution for y^{'''}+4y^'=0 is y=c_{1}+c_{2}cos(2t)+c_{3}sin(2t)