Question

Question #d88e8

Answer 1

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

Explanation:

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`