What is the answer to solvefor a,y=ae^x+be^{3x} ?

The answer to solvefor a,y=ae^x+be^{3x} is a=(y-e^{3x}b)/(e^x)

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solvefor a,y=ae^x+be^{3x}

solve for

a, y = a e^{x} + b e^{3 x}

a = \frac{y - e ^{3 x} b}{e ^{x}}

Solution steps

y = a e^{x} + b e^{3 x}

Switch sides

a e^{x} + b e^{3 x} = y

Move

b e^{3 x}

to the right side

a e^{x} = y - b e^{3 x}

Divide both sides by

e^{x}

a = \frac{y - e ^{3 x} b}{e ^{x}}

4 x - 9 = 11

x, 3 x + y = 17

0.4 + 0.5 z = 0.6 z + 0.1 z + 2

6^{- 3} = \frac{1}{216}

lo g_{4} (64) = 2 x + 1

What is the answer to solvefor a,y=ae^x+be^{3x} ?
The answer to solvefor a,y=ae^x+be^{3x} is a=(y-e^{3x}b)/(e^x)