What is the solution for y^{'''}+y^{''}+3y^'-5y=0 ?

The solution for y^{'''}+y^{''}+3y^'-5y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(2t)+c_{3}sin(2t))

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y^{'''}+y^{''}+3y^'-5y=0

y^{^{'''}} + y^{^{''}} + 3 y^{^{'}} - 5 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) + 3 y^{'} (t) - 5 y = 0

Solve linear ODE:

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

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

\int \frac{sin ( 2 x )}{2 sin ( x )} d x

\int_{4}^{\infty} x e^{- 3 x} d x

\frac{d y}{d x} = 2 x y^{2} + 3 x^{2} y^{2}, y (1) = - 1

\int \frac{4 x + 3}{1 - x ^{2}} d x

\frac{d}{d x} (sin (4 x + 7 x^{4}) + e^{1 - x^{2}})

What is the solution for y^{'''}+y^{''}+3y^'-5y=0 ?
The solution for y^{'''}+y^{''}+3y^'-5y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(2t)+c_{3}sin(2t))