What is the solution for y^{'''}+y^{''}+24y^'-26y=0 ?

The solution for y^{'''}+y^{''}+24y^'-26y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(5t)+c_{3}sin(5t))

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y^{'''}+y^{''}+24y^'-26y=0

y^{^{'''}} + y^{^{''}} + 24 y^{^{'}} - 26 y = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) + 24 y^{'} (t) - 26 y = 0

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{x}{y} e^{x}

\frac{\partial}{\partial x} (sin (π (3 x - 4 y)))

\frac{d}{d x} (\frac{5}{( 6 x + 5 ) ^{2}})

\int_{0}^{8} \frac{1}{64 - t ^{2}} d t

f (x) = - \frac{1}{x ^{2}}

What is the solution for y^{'''}+y^{''}+24y^'-26y=0 ?
The solution for y^{'''}+y^{''}+24y^'-26y=0 is y=c_{1}e^t+e^{-t}(c_{2}cos(5t)+c_{3}sin(5t))