Substitutions

Section 2.7 Substitutions

The third and final technique I will present for first order DEs is substitution. At some level, solving DEs is a more complicated and involved version of doing integrals, since the process is trying to undo the results of differentiation. Substitution is the most common and important technique for solving integrals. It takes complicated integrals and changes their setup to make them more approachable. The technique is exactly the same here: I change the DE with some substitution operation to turn it into something I already know how to do. As with substitution for integrals, I can recognize some typical forms, but many others require creativity and ingenuity to solve. In this section, I’ll introduce two forms which use substitution, just as examples of how the techinque works. Thoughout the wide study of differential equations, a host of different substitutions can occur.

Subsection 2.7.1 Homogeneous DEs

The first substitution is for a class of DEs called homogenous equations. Unfortunately, this is a terrible term: this ‘homogeneous’ has nothing to do with the previous use of the term for linear equations. The word here comes from a different use of the term in algebra. This is deeply confusing and unfortunate, but it is the situation. In any case, a homogenous DE has the following form.

\begin{equation*} \frac{dy}{dt} = f \left( \frac{y}{t} \right) \end{equation*}

The substitution that simplifies these equations is \(v = \frac{y}{t}\text{.}\) The right side of the equation easily turns into \(f(v)\text{,}\) but the transformation of the left side is trickier. Let me do some algebra to try to demonstrate it. First, in the substitution I can solve for \(y\text{.}\) Then I can differentiate this expression in \(t\text{.}\)

\begin{align*} y \amp = tv\\ \frac{d}{dt} y \amp = \frac{d}{dt} tv = v + t \frac{dv}{dt} \end{align*}

Now I have an expression that I can use to replace the left side. Changing both the left and the right sides gives a new differential equation in the new function \(v(t)\text{.}\)

\begin{equation*} v + t \frac{dv}{dt} = f(v) \end{equation*}

Now I solve for \(\frac{dv}{dt}\) to put this new DE in a conventional form.

\begin{equation*} \frac{dv}{dt} = \frac{f(v) - v}{t} \end{equation*}

This is now a separable equation. The substitution has changed a potentially strange and difficult equation into one I know how to do from a previous techique.

Example 2.7.1.

This example is already a separable equation, so I don’t really need the technique of substitution. However is it a nice, relatively easy example to show how this kind of substitution works.

\begin{align*} \frac{dy}{dt} \amp = \frac{t}{y}\\ v \amp = \frac{y}{t}\\ f(v) \amp = \frac{1}{v}\\ \frac{dv}{dt} \amp = \frac{\frac{1}{v} - v}{t} \end{align*}

This is the new seperable equation. I’ll solve it in the conventional method.

\begin{align*} \int \frac{v}{1-v^2} dv \amp = \ln |t| + c\\ \frac{-1}{2} \ln |1-v^2| \amp = \ln |t| + C\\ \frac{1}{\sqrt{1-v^2}} \amp = ct\\ \sqrt{1-v^2} \amp = \frac{c}{t}\\ 1-v^2 \amp = \frac{c}{t^2}\\ v^2 \amp = 1 - \frac{c}{t^2}\\ v \amp = \pm \sqrt{ 1 - \frac{c}{t^2}} \end{align*}

After solving the separable equation, now I reverse the substitution and try to solve for the original function \(y\text{.}\)

\begin{align*} \frac{y}{t} \amp = \pm \sqrt{ 1 - \frac{c}{t^2}}\\ y \amp = \pm t \sqrt{1 - \frac{c}{t^2}} = \pm \sqrt{t^2 - c} \end{align*}

Example 2.7.2.

\begin{equation*} \frac{dy}{dt} = \frac{-y^2 - yt }{t^2} = \frac{-y^2}{t^2} - \frac{y}{t} \end{equation*}

This is a homogeneous equation, so I set up the substitution.

\begin{align*} v \amp = \frac{y}{t}\\ f(v) \amp = -v^2 -v\\ \frac{dv}{dt} \amp = \frac{f(v) - v}{t} = \frac{-v^2 -v -v}{t} = \frac{-v^2 - 2v}{t} \end{align*}

Now I have my separable equation, which I proceed to solve. This isn’t the easiest separable equation. On the left, in the \(v\) integral, I need to use partial fractions to break up the rational function into easier pieces.

\begin{align*} \int \frac{1}{-v^2-2v} dv \amp = \int \frac{1}{t} dt = \ln |t| + c\\ \frac{1}{-v^2 - 2v} \amp = \frac{-1}{v(2+v)} = \frac{1}{2} \frac{1}{v+2} + \frac{-1}{2} \frac{1}{v} \end{align*}

After partial fractions, I can proceed with the left integral. After the integral, there is a series of algebraic steps to try to isolate \(v\text{.}\)

\begin{align*} \frac{1}{2} \int \frac{1}{v+2} dv + \frac{-1}{2} \int \frac{1}{v} dv \amp = \ln |t| + c\\ \frac{1}{2} \ln |v+2| + \frac{-1}{2} \ln |v| \amp = \ln |t| + c\\ \ln |v+2|^{\frac{1}{2}} + \ln |v|^\frac{-1}{2} \amp = \ln |t| + c\\ \sqrt{1 + \frac{2}{v}} \amp = ct\\ 1 + \frac{2}{v} \amp = ct^2\\ \frac{2}{v} \amp = ct^2 -1\\ v \amp = \frac{2}{ct^2 -1} \end{align*}

Finally, after \(v\) is isolated, the substitution can be reversed to find the final answer.

\begin{align*} \frac{y}{t} \amp = \frac{2}{ct^2 - 1}\\ y \amp = \frac{2t}{ct^2 -1} \end{align*}

Subsection 2.7.2 Bernoulli Equations

The other substitution I want to present is for a class of DEs called Bernoulli equations. These equations are nearly linear, but they have an extra \(y^n\) term. They are written in a form very similar to the form for linear DEs.

\begin{equation*} \frac{dy}{dt} + P(t) y = Q(t) y^n\text{.} \end{equation*}

The substitution that simplifies a Bernoulli equation is \(v = y^{1-n}\text{.}\) Be careful: this is \(y^{1-n}\text{,}\) not \(y^{n-1}\text{.}\) It’s a very easy mistake to get this exponent wrong. I transform the DE by looking at the following derivations. The first line is a chain rule. After the chain rule, I use the original DE to replace \(\frac{dy}{dt}\) with \((Qy^n - P)\text{.}\)

\begin{align*} \frac{dv}{dt} \amp = (1-n) y^{1-n-1} \frac{dy}{dt} = (1-n)y^{-n} \frac{dy}{dt}\\ \amp = (1-n)y^{-n} \left( Q y^n - P y \right) \\ \amp = (1-n) Q - (1-n) P y^{1-n}\\ \amp = (1-n) Q - (1-n) P v\\ \frac{dv}{dt} + (1-n)v P \amp = (1-n) Q \end{align*}

This is now a linear equation in \(v\text{.}\) I can solve it as a linear equation in \(v\) and then use the reverse substitution to get back to \(y\text{.}\)

Example 2.7.3.

\begin{align*} \frac{dy}{dt} - \frac{1}{2} \frac{y}{t} \amp = -e^t y^3\\ n \amp =3\\ v \amp = y^{-2}\\ \frac{dv}{dt} - (-2) v \frac{1}{2t} \amp = -2 (-e^t) = 2e^t\\ \frac{dv}{dt} + \frac{v}{t} \amp = 2e^t \end{align*}

This is a familiar linear equation which was solved in Example 2.6.4 Instead of repeating that solution, I’ll use the result from the previous example. After solving the linear DE, I reverse the substitution and solve for \(y\) to get the final result.

\begin{align*} v \amp = 2e^t \left( 1 + \frac{1}{t} \right) + \frac{c}{t}\\ y^{-2} \amp = 2e^t \left( 1 + \frac{1}{t} \right) + \frac{c}{t}\\ y \amp = \left( 2e^t \left( 1 + \frac{1}{t} \right) + \frac{c}{t} \right)^{\frac{-1}{2}}\\ y \amp = \frac{\pm1}{\sqrt{2e^t \left( 1 + \frac{1}{t} \right) + \frac{c}{t}}} \end{align*}

Example 2.7.4.

\begin{align*} \frac{dy}{dt} \amp = y(ty^3 -1) = -y + ty^4 \\ \frac{dy}{dt} + y \amp = ty^4 \\ n = 4\\ v = y^{1-4} = y^{-3} \\ P(t) \amp = 1\\ Q(t) \amp = t \end{align*}

These are the setup pieces. I can now write the associated linear DE, following the form I derive in general.

\begin{align*} \frac{dv}{dt} - 3v \amp = -3t \end{align*}

Now I solve the linear DE in the standard method, by calculating the integrating factor, multiplying by the integrating factor and treating the left side of the equation as a product rule derivative.

\begin{align*} \mu\amp = e^{\int Pdt} = e^{-3t}\\ \frac{d}{dt} e^{-3t} v \amp = -3t e^{-3t}\\ e^{-3t} v \amp = \int -3t e^{-3t} dt = -3 \left( \frac{te^{-3t}}{-3} - \int \frac{e^{-3t}{-3}} dt \right)\\ e^{-3t} v \amp = t e^{-3t} - \int e^{-3t} dt = t e^{-3t} + \frac{e^{-3t}}{3} + c\\ v = \amp t + \frac{1}{3} + ce^{3t} = \frac{3t + 1 + ce^{3t}}{3} \end{align*}

Now I reverse the substitution and try to solve for the original function.

\begin{align*} y^{-3} \amp = \frac{3t + 1 + ce^{3t}}{3}\\ y^3 \amp = \frac{3}{3t + 1 + ce^{3t}}\\ y \amp = \sqrt[3]{\frac{3}{3t + 1 + ce^{3t}}} \end{align*}

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