is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Substitution Methods

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. $$(x + y + 3)^2 \to v^2 = \dfrac{dv}{dx} - 1$$ $$\dfrac{dv}{dx} = v^2 + 1$$ $$\int\dfrac{1}{v^2 + 1}\, dv = \int 1\, dx$$ $$\arctan{v} = x + C \rightarrow v = \tan(x + C)$$ $$x + y + 3 = \tan(x + C) \to y = \tan(x + C) - x - 3$$

Example Two: v = y/x

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.
David Witten

Exact Equations

Surface Integrals