What is the answer to solvefor a,y=ae^{1/2 (-2)}+be^{-1/2 (-2)} ?

The answer to solvefor a,y=ae^{1/2 (-2)}+be^{-1/2 (-2)} is a=ey-e^2b

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solvefor a,y=ae^{1/2 (-2)}+be^{-1/2 (-2)}

solve for

a, y = a e^{\frac{1}{2} (- 2)} + b e^{- \frac{1}{2} (- 2)}

a = ey - e^{2} b

Solution steps

y = a e^{\frac{1}{2} (- 2)} + b e^{- \frac{1}{2} (- 2)}

Switch sides

a e^{\frac{1}{2} (- 2)} + b e^{- \frac{1}{2} (- 2)} = y

Simplify

a e^{\frac{1}{2} (- 2)} + b e^{- \frac{1}{2} (- 2)} : \frac{1}{e} a + e b

\frac{1}{e} a + e b = y

Move

e b

to the right side

\frac{1}{e} a = y - e b

Multiply both sides by

e

a = ey - e^{2} b

\frac{( 4 - x )}{( - 2 )} \leq 1

m (b - a) - p (a - b)

49 t^{6} - 4 k^{8}

48 = 6 x

\frac{9}{8} \cdot \frac{7}{3} \cdot \frac{5}{14}

What is the answer to solvefor a,y=ae^{1/2 (-2)}+be^{-1/2 (-2)} ?
The answer to solvefor a,y=ae^{1/2 (-2)}+be^{-1/2 (-2)} is a=ey-e^2b