What is the solution for y^{'''}+4y^{''}-5y^'=0 ?

The solution for y^{'''}+4y^{''}-5y^'=0 is y=c_{1}+c_{2}e^t+c_{3}e^{-5t}

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y^{'''}+4y^{''}-5y^'=0

y^{^{'''}} + 4 y^{^{''}} - 5 y^{^{'}} = 0

y = c_{1} + c_{2} e^{t} + c_{3} e^{- 5 t}

Solution steps

y^{'''} (t) + 4 y^{''} (t) - 5 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{t} + c_{3} e^{- 5 t}

y = c_{1} + c_{2} e^{t} + c_{3} e^{- 5 t}

\int_{0}^{8} (8 x - x^{2}) d x

\int (8 x - 12) (4 x^{2} - 12 x)^{4} d x

\frac{d}{d y} (e^{2 x + y})

f (x) = arctan (x)

\frac{d}{d x} (x^{2} \cdot e^{- 4 x})

What is the solution for y^{'''}+4y^{''}-5y^'=0 ?
The solution for y^{'''}+4y^{''}-5y^'=0 is y=c_{1}+c_{2}e^t+c_{3}e^{-5t}