MathJax TeX Test Page
Let's say we're given this differential equation that we can't solve (with our current methods).
$$\dfrac{dy}{dx} = f(x,y)$$
We want to rewrite the equation in terms of a different variable, $v$. So, we do a substitution, like $v = x + y$, although it can be anything. Now, we rewrite it in terms of y,$y = \phi(x,v)$, and take the derivative with respect to x. The next step is done with the multivariable chain rule.
$$\dfrac{dy}{dx} = \phi_x(x,v)\dfrac{dx}{dx} + \phi_v(x,v)\dfrac{dv}{dx} = \boxed{\phi_x(x,v) + \phi_v(x,v)\dfrac{dv}{dx}}$$
So, we set the two expressions for $\frac{dy}{dx}$, and then we're guaranteed to be able to solve the equation. This is because we are only doing a few substitutions, $v = ax + by + c$, $v = \frac{y}{x}$, and $v = y^{1 - n}$. We proved that using each substitution gives you a separable or first-order linear equation.
# Example One: v = ax + by + c

MathJax TeX Test Page
$$\dfrac{dy}{dx} = (x + y + 3)^2$$
Here is a perfect opportunity to substitute.
$v = x + y + 3 \to y = v - x - 3$ simplifies it a lot. Now, we differentiate the **new** expression for y.
$$\dfrac{dy}{dx} = \dfrac{dv}{dx} - 1$$
Now, we set the two expressions equal.
$$(x + y + 3)^2 \to v^2 = \dfrac{dv}{dx} - 1$$
$$\dfrac{dv}{dx} = v^2 + 1$$
$$\int\dfrac{1}{v^2 + 1}\, dv = \int 1\, dx$$
$$\arctan{v} = x + C \rightarrow v = \tan(x + C)$$
$$x + y + 3 = \tan(x + C) \to y = \tan(x + C) - x - 3$$

MathJax TeX Test Page
$$2xy\dfrac{dy}{dx} = 4x^2 + 3y^2$$
We want $\dfrac{dy}{dx}$ on its own side.
$$\dfrac{dy}{dx} = \dfrac{4x^2 + 3y^2}{2xy} = 2\left(\dfrac{x}{y}\right) + \frac{3}{2}\left(\dfrac{y}{x}\right)$$
Here, we can see $\frac{y}{x}$ looks like an appropriate substitution.
$$v = \frac{y}{x} \to y = vx$$
Remember the multivariable chain rule:$ \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$.
$$\dfrac{dy}{dx} = v\frac{dx}{dx} + x\frac{dv}{dx} = v + x\dfrac{dv}{dx}$$
Now, we set the two equations for $\frac{dy}{dx}$ equal.
$$\dfrac{2}{v} + \dfrac{3}{2}v = v + x\dfrac{dv}{dx}$$
$$\dfrac{2}{v} + \dfrac{v}{2} \to \dfrac{v^2 + 4}{2v} = x\dfrac{dv}{dx}$$
$$\int \dfrac{2v}{v^2 + 4}\, dv = \int \dfrac{1}{x} \, dx$$
$$\ln|v^2 + 4| = \ln|x| + C \to v^2 + 4 = C_2x$$
Note that this is a different constant. It is initially positive because it's equal to $e^c$, but it absorbs the absolute value.
$$y^2 + 4x^2 = Cx^3$$
# Example Three: v = y^(1 - n)

MathJax TeX Test Page
These problems look like $\frac{dy}{dx} + P(x)y = Q(x)y^n$.
Let's look at our last problem again.
$$2xy\dfrac{dy}{dx} = 4x^2 + 3y^2$$
Once again, we $\frac{dy}{dx}$ to be on its own side, so we can eventually set it equal to its other equation in terms of v.
$$\dfrac{dy}{dx} = \dfrac{3}{2x}y + \dfrac{2x}{y}$$
Now, however, we want it to look like $\frac{dy}{dx} + P(x)y = Q(x)y^n$, so we can perform the substitution.
$$\dfrac{dy}{dx} - \dfrac{3}{2x}y = \dfrac{2x}{y}$$
In this problem $y^n = y^{-1}$, so we substitute $v = y^{1 - (-1)} = y^2 \to y = v^{\frac{1}{2}}.$ Now, we set the two expressions equal.
$$\dfrac{dy}{dx} = \frac{1}{2}v^{-1/2}\frac{dv}{dx} = \frac{3}{2x}v^{1/2} + 2xv^{-1/2}$$
$$\dfrac{dv}{dx} - \frac{3}{x}v = 4x$$
Now, we can solve this using an integrating factor. I'll write a post about that.