Question

(x^2 + y^2) dx + (3xy) dy = 0
This equation isn't exact. I am wondering if I should use the equations for a special integrating factor.

Answer 1

you can use formulas for bernoulli equations:
http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx

rearranging gives:
\[ \frac{dy}{dx} + \frac{y}{3x} = -\frac{xy^{-1}}{3}\]
\[y \frac{dy}{dx} + \frac{y^2}{3x} = -\frac{x}{3}\]
make the substitution
v = y^2
v' = 2y(dy/dx)
\[\rightarrow \frac{v'}{2} + \frac{v}{3x} = -\frac{x}{3}\]
now use the integrating factor x^(2/3)
\[\rightarrow (x^{2/3} v)' = -\frac{2}{3}x^{5/3}\]
integrating
\[x^{2/3} v = -\frac{1}{4} x^{8/3}+C\]
\[v = -\frac{x^2}{4}+Cx^{-2/3}\]
plug "y" back in ..... v=y^2
\[y = \pm \sqrt{Cx^{-2/3}-\frac{x^2}{4}}\]

Answer 2

Never heard of that and I don't think that is what my prof is looking for

Answer 3

Calculate the vapor pressure at 20 ∘C of a saturated solution of the nonvolatile solute, urea, CO(NH2)2 in methanol, CH3OH. The solubility is 17 g urea/100 mL methanol. The density of methanol is 0.792 g/mL, and its vapor pressure at 20 ∘C is 95.7 mmHg.