eliminate the arbitrary constant from y= Ae^2x +Be^x + C



Answer to a math question eliminate the arbitrary constant from y= Ae^2x +Be^x + C

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To eliminate the arbitrary constant from the given equation, we need to find the relationship between A, B, and C. We can do this by differentiating the equation with respect to x.

Step 1: Differentiate the equation y = Ae^(2x) + Be^x + C with respect to x.

\frac{dy}{dx} &= \frac{d}{dx}(Ae^(2x)) + \frac{d}{dx}(Be^x) + \frac{d}{dx}(C) \
&= 2Ae^(2x) + Be^x + 0 \
\frac{dy}{dx} &= 2Ae^(2x) + Be^x

Step 2: Now, equate this differentiated equation to zero to find the relationship between A, B, and C.

2Ae^(2x) + Be^x = 0

Step 3: Factor out the common exponential term e^x.

e^x (2Ae^x + B) = 0

Step 4: Since e^x is always positive and never equal to zero, we can set the expression within parentheses equal to zero.

2Ae^x + B = 0

Step 5: Solve this equation for A in terms of B.

2Ae^x = -B

Ae^x = -B/2

A = -\frac{B}{2e^x}

Step 6: Substitute the expression for A back into the original equation y = Ae^(2x) + Be^x + C.

y = (-\frac{B}{2e^x})e^(2x) + Be^x + C

y = -\frac{Be^x}{2} + Be^x + C

Step 7: Simplify the expression by finding a common denominator.

y = \frac{2Be^x - Be^x}{2} + C

y = \frac{Be^x}{2} + C

Answer: The equation without the arbitrary constant is y = \frac{Be^x}{2} + C.

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