Given that y=3e2x+2e−xy = 3e^{2x} + 2e^{-x}, and that d2ydx2+dydx+y=Ae2x+Be−x\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = Ae^{2x} + Be^{-x}, find the values of each of the constants AA and BB.

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Worked solution

2

1. Find the first derivative of $y$ with respect to $x$.

Given y=3e2x+2e−xy = 3e^{2x} + 2e^{-x},
dydx=3(2)e2x+2(−1)e−x\frac{\text{d}y}{\text{d}x} = 3(2)e^{2x} + 2(-1)e^{-x}
dydx=6e2x−2e−x\frac{\text{d}y}{\text{d}x} = 6e^{2x} - 2e^{-x}

2. Find the second derivative of $y$ with respect to $x$.

d2ydx2=6(2)e2x−2(−1)e−x\frac{\text{d}^2 y}{\text{d}x^2} = 6(2)e^{2x} - 2(-1)e^{-x}
d2ydx2=12e2x+2e−x\frac{\text{d}^2 y}{\text{d}x^2} = 12e^{2x} + 2e^{-x}

3. Substitute $y$, $\frac{\text{d}y}{\text{d}x}$, and $\frac{\text{d}^2 y}{\text{d}x^2}$ into the given expression.

d2ydx2+dydx+y=(12e2x+2e−x)+(6e2x−2e−x)+(3e2x+2e−x)\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = (12e^{2x} + 2e^{-x}) + (6e^{2x} - 2e^{-x}) + (3e^{2x} + 2e^{-x})

4. Simplify the expression by collecting like terms.

d2ydx2+dydx+y=(12+6+3)e2x+(2−2+2)e−x\frac{\text{d}^2 y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} + y = (12 + 6 + 3)e^{2x} + (2 - 2 + 2)e^{-x}
=21e2x+2e−x= 21e^{2x} + 2e^{-x}

5. Compare coefficients to find $A$ and $B$.

Comparing 21e2x+2e−x21e^{2x} + 2e^{-x} with Ae2x+Be−xAe^{2x} + Be^{-x}:
A=21A = 21
B=2B = 2

Answer

A=21A = 21, B=2B = 2

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