Linear Substitution

If a first order equation isn’t separable or linear, then one of three types of substitutions might solve the equation.

The Essentials

The first kind of substitution is a linear substitution:

\[ u = ax + by + c \]

This substitution can be used if the equation is in the form:

\[ \frac{dy}{dx} = f(ax + by + c) = f(x, y) \]

The substitution is solved for y in terms of u and x, and then it is differentiated to find \( \frac{dy}{dx} \):

\[ y = \frac{u - ax - c}{b} \] \[ \frac{dy}{dx} = \frac{1}{b} \left( \frac{du}{dx} - a \right) \]

Then in the original equation \( \frac{dy}{dx} \) is switched in as well as u:

\[ \frac{dy}{dx} = f(ax + by + c) \] \[ \frac{1}{b} \left( \frac{du}{dx} - a \right) = f(u) \]

Example

Solve the equation:

\[ \frac{dy}{dx} = y - 2x - 1 + \frac{1}{2x - y + 3} \]

This equation can be solved by using the substitution for either linear term: \(u = y−2x−1 \) or \(u = 2x−y+3 \). In this example we will just show the process with the second:

\[ u = 2x - y + 3 \] \[ y = 2x - u + 3 \] \[ \frac{dy}{dx} = -\frac{du}{dx} + 2 \]

The equation can be re-written with the substitution:

\[ \frac{dy}{dx} = y - 2x - 1 + \frac{1}{2x - y + 3} \] \[ -\frac{du}{dx} + 2 = -u + 2 + \frac{1}{u} \] \[ \frac{du}{dx} = u - \frac{1}{u} \] \[ \frac{du}{dx} = \frac{u^2 - 1}{u} \]

This is a separable differential equation. Next, we solve the equation and undo the substitution:

\[ \frac{u}{u^2 - 1} \frac{du}{dx} = 1 \] \[ \int \frac{u}{u^2 - 1} \, du = \int \, dx \] \[ \frac{1}{2} \ln |u^2 - 1| = x + C \] \[ \ln |u^2 - 1| = 2x + C \] \[ u^2 - 1 = Ce^{2x} \] \[ u^2 = Ce^{2x} + 1 \] \[ (2x - y + 3)^2 = Ce^{2x} + 1 \] \[ 2x - y + 3 = \pm \sqrt{Ce^{2x} + 1} \] \[ y(x) = 2x + 3 \mp \sqrt{Ce^{2x} + 1} \]

Practice

Evaluate this expressions:

\[ \sin(-6y + 2x + 7) \frac{dy}{dx} = 3 + 3 \sin(-6y + 2x + 7) \]

Solution:

\[ y(x) = 3x + \frac{2}{7} - \frac{1}{2} \cos^{-1}(6x + C) \]