What is the solution for y^{(5)}+9y^'=0 ?

The solution for y^{(5)}+9y^'=0 is

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y^{(5)}+9y^'=0

y^{(5)} + 9 y^{^{'}} = 0

y = c_{1} + e^{- \frac{3}{2} t} (c_{2} cos (\frac{3}{2} t) + c_{3} sin (\frac{3}{2} t)) + e^{\frac{3}{2} t} (c_{4} cos (\frac{3}{2} t) + c_{5} sin (\frac{3}{2} t))

Solution steps

y^{(5)} + 9 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + e^{- \frac{3}{2} t} (c_{2} cos (\frac{3}{2} t) + c_{3} sin (\frac{3}{2} t)) + e^{\frac{3}{2} t} (c_{4} cos (\frac{3}{2} t) + c_{5} sin (\frac{3}{2} t))

y = c_{1} + e^{- \frac{3}{2} t} (c_{2} cos (\frac{3}{2} t) + c_{3} sin (\frac{3}{2} t)) + e^{\frac{3}{2} t} (c_{4} cos (\frac{3}{2} t) + c_{5} sin (\frac{3}{2} t))

x (y^{6} + 1) d x + y^{2} (x^{4} + 1) d y = 0

\frac{d y}{d x} - 2 y = 6, y = 2, x = 0

(1 + e^{x}) y \cdot y^{^{'}} = e^{x}

\frac{d y}{d x} = \frac{10 ( 1 + x )}{x y}

x^{^{'}} = e^{x} t cos (t)