What is the solution for y^{''}+2y^'-48y=0 ?

The solution for y^{''}+2y^'-48y=0 is y=c_{1}e^{6t}+c_{2}e^{-8t}

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y^{''}+2y^'-48y=0

y^{^{''}} + 2 y^{^{'}} - 48 y = 0

y = c_{1} e^{6 t} + c_{2} e^{- 8 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 48 y = 0

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{- 8 t}

y = c_{1} e^{6 t} + c_{2} e^{- 8 t}

\int \frac{u}{u} d u

f (x) = \frac{1}{x} + 4, (- 2, 8)

\frac{\partial}{\partial x} (3 t^{2} e^{4 x})

x \to 1 lim ((3 x - 1)^{2})

x^{2} y^{^{''}} + 3 x y^{^{'}} + y = 0

What is the solution for y^{''}+2y^'-48y=0 ?
The solution for y^{''}+2y^'-48y=0 is y=c_{1}e^{6t}+c_{2}e^{-8t}