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

The answer to solvefor a,z=ae^{3t}+be^t is a=(z-e^tb)/(e^{3t)}

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solvefor a,z=ae^{3t}+be^t

solve for

a, z = a e^{3 t} + b e^{t}

a = \frac{z - e ^{t} b}{e ^{3 t}}

Solution steps

z = a e^{3 t} + b e^{t}

Switch sides

a e^{3 t} + b e^{t} = z

Move

b e^{t}

to the right side

a e^{3 t} = z - b e^{t}

Divide both sides by

e^{3 t}

a = \frac{z - e ^{t} b}{e ^{3 t}}

29 w - 18 = 40

3 (4 x - 5) + 2 (11 - 2 x) = 43

\frac{x}{- 6} = 10

\frac{1}{3} (d + 9) = - 12

10 = 8 + s

What is the answer to solvefor a,z=ae^{3t}+be^t ?
The answer to solvefor a,z=ae^{3t}+be^t is a=(z-e^tb)/(e^{3t)}