A Differential Equation

The differential equation is of the form $\frac{dy}{dx} + yP(x) = Q(x)$ and therefore the integrating factor method can be used. Let the integrating factor be $I(x)$ .

$I(x) = e^{\int P(x) \, dx} = e^{\int{\ln{x}} \, dx} = e^{x(\ln{x} - 1)} = x^{x} e^{-x}$

Multiplying each side of the differential equation by the integrating factor

$\frac{dy}{dx} x^{x} e^{-x} + y x^{x} e^{-x} \ln{x} = e^{-2x}$

$\frac{d}{dx} y x^{x} e^{-x} = e^{-2x}$

$\int \frac{d}{dx} y x^{x} e^{-x} \, dx = \int e^{-2x} \, dx$

$y x^{x} e^{-x} = -\frac{1}{2} e^{-2x} + c$

$y = \frac{- e^{-2x} + 2c}{2x^{x} e^{-x}}$

Given that $y(1) = 0$

$0 = - e^{-2(1)} + 2c$

$c = \frac{1}{2e^{2}}$

Therefore

$y = \frac{- e^{-2x} + \frac{1}{e^{2}}}{2x^{x} e^{-x}}$

and

$y(2) = \frac{- e^{-2(2)} + \frac{1}{e^{2}}}{2(2)^{(2)} e^{-(2)}} = \frac{e^{2}-1}{8e^{2}} = \frac{e^{2}+a}{be^{2}}$

where

$a = -1$ and $b = 8$

Therefore

$a + b = \boxed{7}$

1) The differential equation is a Bernoulli differential equation, with n = 0, p(x) = ln x, q(x) = (x e)^-x.

2) Using the formula for the Bernoulli differential equation the general equation is: y = [(-1/2) exp (-2x) + C] / [(x^x)(e^-x)].

3) If x = 1, y = 0, then C = (1/2) exp (-2). Therefore the particular becomes: y = [(-1/2) exp (-2x) + (1/2) exp (-2)] / [(x^x)(e^-x)].

4) If x = 2, then y = [(-1/2) exp (-4) + (1/2) exp (-2)] / {2^2)[exp (-2)]}, or y = [ exp(2) - 1] / [8 exp(2)].

5) Equating coefficients: a = -1, b = 8. Therefore, a + b = 7. (Q.E.D.)