# Differential equations help with substitution

• Sep 2nd 2010, 09:22 PM
acevipa
Differential equations help with substitution
Solve:

$\displaystyle y'=\dfrac{2}{(x+2y-3)}$ by letting $\displaystyle x+2y-3=v$

I always get confused up as to what we're meant to do with the $\displaystyle y'$

$\displaystyle y'=\dfrac{2}{v}$

Is $\displaystyle \dfrac{dv}{dx}=1+2\dfrac{dy}{dx}$?
• Sep 2nd 2010, 11:21 PM
Prove It
If $\displaystyle x + 2y - 3 = v$

$\displaystyle 2y = v - x + 3$

$\displaystyle y = \frac{1}{2}v - \frac{1}{2}x + \frac{3}{2}$

$\displaystyle \frac{d}{dx}(y) = \frac{d}{dx}\left(\frac{1}{2}v - \frac{1}{2}x + \frac{3}{2}\right)$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}\,\frac{d}{dx}(v) - \frac{1}{2}\,\frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{3}{2}\right)$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}\,\frac{d}{dv}(v)\,\frac{dv}{dx} - \frac{1}{2}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}\,\frac{dv}{dx} - \frac{1}{2}$.

So substituting into your DE

$\displaystyle \frac{dy}{dx} = \frac{2}{x + 2y - 3}$

$\displaystyle \frac{1}{2}\,\frac{dv}{dx} - \frac{1}{2} = \frac{2}{v}$

$\displaystyle \frac{dv}{dx} - 1 = \frac{4}{v}$

$\displaystyle \frac{dv}{dx} = \frac{4}{v} + 1$

$\displaystyle \frac{dv}{dx} = \frac{v + 4}{v}$

$\displaystyle \frac{dx}{dv} = \frac{v}{v + 4}$

$\displaystyle \frac{dx}{dv} = \frac{v + 4}{v + 4} - \frac{4}{v + 4}$

$\displaystyle \frac{dx}{dv} = 1 - \frac{4}{v + 4}$

$\displaystyle x = \int{1 - \frac{4}{v + 4}\,dv}$

$\displaystyle x = v - 4\ln{|v + 4|} + C$

$\displaystyle x = x + 2y - 3 - 4\ln{|x + 2y - 3 + 4|} + C$

$\displaystyle x = x + 2y - 3 - 4\ln{|x + 2y + 1|} + C$

$\displaystyle 4\ln{|x+2y+1|} = 2y - 3 + C$

$\displaystyle \ln{|x+2y+1|}= \frac{1}{2}y - \frac{3}{4} + C$

$\displaystyle |x + 2y + 1| = e^{\frac{1}{2}y - \frac{3}{4} + C}$

$\displaystyle |x + 2y + 1| = e^{-\frac{3}{4} + C}e^{\frac{1}{2}y}$

$\displaystyle x + 2y + 1 = \pm e^{-\frac{3}{4} + C}e^{\frac{1}{2}y}$

$\displaystyle x + 2y + 1 = Ae^{\frac{1}{2}y}$

$\displaystyle x = Ae^{\frac{1}{2}y} - 2y - 1$.