# Thread: Substitution method

1. ## Substitution 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?

2. ## Re: 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|.

3. ## Re: Substitution method

Originally Posted by alexander9408
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?
\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.