What is the solution for y^{''}-8y=0 ?

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

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y^{''}-8y=0

y^{^{''}} - 8 y = 0

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

Solution steps

y^{''} (t) - 8 y = 0

Solve linear ODE:

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

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

e^{x + y}

\int_{1}^{4} 5 x + 3 x^{2} - 8 d x

\frac{d y}{d x} = - \frac{( 2 y ^{2} )}{x ^{2}}

a x^{^{''}} + b x = 0

\frac{d}{d x} (e^{t a n h (2 x)})

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