Given that the differential equation dy/dx = (x+y+2)/(x+y+1), by using a suitable substitution to reduce it to a differential equation with separable variable.

How do I get a suitable substitution for this? Any method can be used?

Printable View

- Apr 22nd 2013, 08:05 PMalexander9408Substitution method
Given that the differential equation dy/dx = (x+y+2)/(x+y+1), by using a suitable substitution to reduce it to a differential equation with separable variable.

How do I get a suitable substitution for this? Any method can be used? - Apr 23rd 2013, 04:30 AMchiroRe: Substitution method
Hey alexander9408.

You should first try using the fact x + y + 2 = x + y + 1 + 1 so you will get 1 + 1/(x+y+1).

I'm not sure what kind of DE this is though. If you need an analytic result or can you use a numerical scheme? The DE should be very close to f(x) = x since (x+y+2) is almost equal to (x+y+1) for large enough |x+y|. - Apr 23rd 2013, 04:51 AMProve ItRe: Substitution method
$\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{x + y + 2}{x + y + 1} \end{align*}$

Now make the substitution $\displaystyle \displaystyle \begin{align*} v = x + y \implies y = v - x \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \end{align*}$ and the integral becomes

$\displaystyle \displaystyle \begin{align*} \frac{dv}{dx} - 1 &= \frac{v + 2}{v + 1} \\ \frac{dv}{dx} &= \frac{v+2}{v+1} + 1 \\ \frac{dv}{dx} &= \frac{2v + 3}{v + 1} \\ \left( \frac{v + 1}{2v + 3} \right) \frac{dv}{dx} &= 1 \\ \int{\left( \frac{v + 1}{2v +3} \right) \frac{dv}{dx} \, dx} &= \int{1\,dx} \\ \int{\frac{v+1}{2v+3} \, dv} &= \int{1\,dx} \\ \frac{1}{2}\int{\frac{2v+2}{2v+3}\,dv} &= \int{1\,dx} \\ \frac{1}{2}\int{1 - \frac{1}{2v+3}\,dv} &= \int{1\,dx} \end{align*}$

Go from here.