Question #5a53b

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Answer 1 · 2017-03-24T00:16:04

Looks like you're nailing it. Here's how I would have a crack at this.

$xy + y' = 100x$

If we play with the algebra first:

$y'= x (100 - y)$

This is separable:

$(y')/ (100 - y) = x$

So we integrate both sides wrt x:

$int 1/ (100 - y) \ dy/dx \dx = int x \ dx$

By Chain Rule:
$implies int 1/ (100 - y) \dy = int x \ dx$

And so:

$-ln (100 - y) =x^2/2 + C$

$ln (100 - y)^(-1) =x^2/2 + C$

Take each term as an exponential:

$e^( ln (100 - y)^(-1) ) = e^ ( x^2/2 + C)$

$(100 - y)^(-1) = e^C e^ ( x^2/2) = C e^ ( x^2/2)$

$100 - y = C e^ ( -x^2/2)$

$y = 100 - C e^ ( -x^2/2)$