What is the solution for 7y^{''}+7y^'-14y=0 ?

The solution for 7y^{''}+7y^'-14y=0 is y=c_{1}e^t+c_{2}e^{-2t}

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7y^{''}+7y^'-14y=0

7 y^{^{''}} + 7 y^{^{'}} - 14 y = 0

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

Solution steps

7 y^{''} (t) + 7 y^{'} (t) - 14 y = 0

Solve linear ODE:

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

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

7 (3 x^{2} ln (5 x) + x^{2})

x \to 2 + lim (8 - x^{3})

\frac{d}{d t} (3 t cos (t))

10 sin (x)

y = x, y = - x, x = 1, x = 4

What is the solution for 7y^{''}+7y^'-14y=0 ?
The solution for 7y^{''}+7y^'-14y=0 is y=c_{1}e^t+c_{2}e^{-2t}