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Answer 1 · 2017-01-19T12:32:20

$y (x) = a - \frac{c}{| x |}$ $y(x)= a-c/abs(x)$

In the equation:

$x \frac{d y}{d x} + y = a$ $x(dy)/(dx) + y = a$

we can separate the variables in this way:

$x \frac{d y}{d x} = a - y$ $x(dy)/(dx) = a-y$

$\frac{d y}{a - y} = \frac{d x}{x}$ $(dy)/(a-y) = (dx)/x$

then integrate both sides:

$\int \frac{d y}{a - y} = \int \frac{d x}{x}$ $int (dy)/(a-y) = int (dx)/x$

$- \int \frac{d (a - y)}{a - y} = \int \frac{d x}{x}$ $-int (d(a-y))/(a-y) = int (dx)/x$

$- ln | a - y | = ln | x | + C$ $-lnabs(a-y) = lnabs(x) + C$

$ln | a - y | = - ln | x | + C$ $lnabs(a-y) = -lnabs(x) + C$

We can now take the exponential of both sides posing $c = e^{C} > 0$ $c=e^C > 0$ :

$| a - y | = e^{- ln | x | + C} = \frac{e^{C}}{e^{ln | x |}} = \frac{c}{| x |}$ $abs(a-y) = e^(-lnabs(x) + C) = e^C/ e^(lnabs(x)) = c/abs(x)$

If we make the hypothesis $y < a$ $y<a$ we have:

$a - y = \frac{c}{| x |}$ $a-y = c/abs(x)$

$y = a - \frac{c}{| x |}$ $y= a-c/abs(x)$

which is in fact always less than $a$ $a$