What is the solution for y^{''}+2y^'-35y=0 ?

The solution for y^{''}+2y^'-35y=0 is y=c_{1}e^{5t}+c_{2}e^{-7t}

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y^{''}+2y^'-35y=0

y^{^{''}} + 2 y^{^{'}} - 35 y = 0

y = c_{1} e^{5 t} + c_{2} e^{- 7 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 35 y = 0

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 7 t}

y = c_{1} e^{5 t} + c_{2} e^{- 7 t}

x \to 5 - lim (\frac{8}{x - 5})

y = 6^{- 4 x}

\frac{d}{d t} (e^{t} t)

\frac{d}{d x} ((2 + 3 x)^{3})

e^{t}

What is the solution for y^{''}+2y^'-35y=0 ?
The solution for y^{''}+2y^'-35y=0 is y=c_{1}e^{5t}+c_{2}e^{-7t}