# Question #5a53b

Mar 24, 2017

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

#### Explanation:

$x y + y ' = 100 x$

If we play with the algebra first:

$y ' = x \left(100 - y\right)$

This is separable:

$\frac{y '}{100 - y} = x$

So we integrate both sides wrt x:

$\int \frac{1}{100 - y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} \setminus \mathrm{dx} = \int x \setminus \mathrm{dx}$

By Chain Rule:
$\implies \int \frac{1}{100 - y} \setminus \mathrm{dy} = \int x \setminus \mathrm{dx}$

And so:

$- \ln \left(100 - y\right) = {x}^{2} / 2 + C$

$\ln {\left(100 - y\right)}^{- 1} = {x}^{2} / 2 + C$

Take each term as an exponential:

${e}^{\ln {\left(100 - y\right)}^{- 1}} = {e}^{{x}^{2} / 2 + C}$

${\left(100 - y\right)}^{- 1} = {e}^{C} {e}^{{x}^{2} / 2} = C {e}^{{x}^{2} / 2}$

$100 - y = C {e}^{- {x}^{2} / 2}$

$y = 100 - C {e}^{- {x}^{2} / 2}$