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Answer 1 · 2017-01-19T12:32:20

$y(x)= a-c/abs(x)$

In the equation:

$x(dy)/(dx) + y = a$

we can separate the variables in this way:

$x(dy)/(dx) = a-y$

$(dy)/(a-y) = (dx)/x$

then integrate both sides:

$int (dy)/(a-y) = int (dx)/x$

$-int (d(a-y))/(a-y) = int (dx)/x$

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

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

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

$abs(a-y) = e^(-lnabs(x) + C) = e^C/ e^(lnabs(x)) = c/abs(x)$

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

$a-y = c/abs(x)$

$y= a-c/abs(x)$

which is in fact always less than $a$