Simplification by Substitution

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So far in this chapter — as well as in Chapter ?? — we have seen how to find solutions of ordinary differential equations that have special forms. (For instance, separation of variables can be applied to solve equations of the form $\frac {dx}{dt}=g(x) h(t)$ .) However, “most” differential equations are not of the form needed to apply one of these techniques. But sometimes it is possible to transform the equation into such a form. This is accomplished by substituting a new function for $x(t)$ , so that the differential equation for this new function has a simpler form. We illustrate this procedure in two cases.

Homogeneous Coefficients

Differential equations of the form

$\begin{equation} \label{eq:homcoeff} \frac{dx}{dt} = F\left(\frac{x}{t}\right), \end{equation}$ where $F:\R \to \R$ is continuous, are said to have homogeneous coefficients. In general, none of the techniques described so far can be applied to this equation.

Since the function $x(t)/t$ appears as the argument of $F$ in the equation it seems plausible to write down (??) in terms of this function. Having this in mind, define $v(t) = \frac {x(t)}{t}.$ Then $x(t) = t v(t)$ and (??) becomes $v(t) + t \frac {dv}{dt}(t) = F(v(t)).$ When $t\not = 0$ , this equation is equivalent to

$\begin{equation} \label{eq:homsep} \frac{dv}{dt} = \frac{F(v)-v}{t}. \end{equation}$ We have arrived at an equation to which we can apply separation of variables. Once a solution $v(t)$ of (??) is found, we obtain a solution of (??) by setting $x(t) = t v(t).$ If an initial condition $x(t_0)=x_0$ is specified in (??), then we have to solve (??) with the transformed initial condition $v(t_0)=x_0/t_0$ .

An Example

Consider the initial value problem

$\begin{equation} \label{eq:homcoex1} \begin{array}{rcl} \dps \frac{dx}{dt} & = & \dps \frac{2t^2+x^2}{tx} \\ x(2) & = & 6. \end{array} \end{equation}$ Since the right hand side of this equation can be written as $F\left (\frac {x}{t}\right ) = 2\frac {t}{x}+\frac {x}{t},$ we see that (??) has homogeneous coefficients. We set $v(t) = x(t)/t$ . By (??) we have to solve the initial value problem $\begin {array}{rcl} \dps \frac {dv}{dt} & = & \dps \frac {F(v)-v}{t} = \frac {2\frac {1}{v}+v-v}{t}=\frac {2}{tv} \\ v(2) & = & 6/2 = 3. \end {array}$ Separation of variables shows that $v(t)$ has to satisfy $\frac {v^2}{2} = \ln (t) - \ln (2) + \frac {9}{2}$ (see Section ??). Therefore $v(t) = \sqrt {9+2\ln \frac {t}{2}}.$ Finally, we obtain the solution $x(t) = tv(t) = t\sqrt {9+2\ln \frac {t}{2}}.$

Bernoulli’s Equation

Let $r,s:\R \rightarrow \R$ be continuous functions, and let $p\in \R$ be a real number. Then an equation of the form

$\begin{equation} \label{eq:bern1} \frac{dx}{dt} = r(t)x + s(t)x^p \end{equation}$ is called a Bernoulli equation. For $p=0$ or $p=1$ the equation is linear and we can, in principle, find all the solutions by variation of parameters (see Chapter ??). Hence we assume that $p\not = 0,1.$ The idea is to substitute $x(t)$ in such a way that also for these values of $p$ (??) becomes linear in the new function $v(t)$ . We try the guess $v(t) = x(t)^{1/\alpha },$ where we choose the real constant $\alpha \not = 0$ later to simplify the transformed equation. Using the chain rule on $x(t) = v(t)^\alpha$ , we compute $\frac {dx}{dt} = \alpha v^{\alpha -1} \frac {dv}{dt}.$ Substitution into (??) yields $\alpha v^{\alpha -1} \frac {dv}{dt} = r(t)v^\alpha + s(t)v^{p\alpha }.$ Thus $\frac {dv}{dt} = \frac {1}{\alpha } r(t) v + \frac {1}{\alpha } s(t) v^{p\alpha -\alpha +1}.$ To simplify this equation, choose the constant $\alpha$ so that $p\alpha -\alpha +1=0$ ; that is, set $\alpha = \frac {1}{1-p}.$ Then we arrive at the linear equation $\begin{equation} \label{eq:bern2} \frac{dv}{dt} = (1-p) r(t) v + (1-p) s(t), \end{equation}$ which we can, in principle, be solved by variation of parameters. If this can be done, then we obtain a solution $x(t)$ of (??) by setting $x(t) = (v(t))^{\frac {1}{1-p}}.$ If an initial condition $x(t_0)=x_0$ is specified in (??), then we have to solve (??) with the transformed initial condition $v(t_0)=x_0^{1/\alpha }=x_0^{1-p}$ .

An Example

Consider the initial value problem $\begin {array}{rcl} \dps \frac {dx}{dt} & = & 3t^2 x + 3t^2x^{\frac {2}{3}}\\ x(0) & = & 27. \end {array}$ The differential equation is a Bernoulli equation with $r(t) = 3t^2,\quad s(t)=3t^2, \AND p=\frac {2}{3}.$ Hence we have to find a solution of the linear initial value problem (see (??)) $\begin {array}{rcl} \dps \frac {dv}{dt} & = & t^2 v + t^2 \\ v(0) & = & 27^{1/3} = 3. \end {array}$ The solution to this differential equation can be obtained by variation of parameters and is $v(t) = -1 +4 e^{t^3/3}.$ Therefore, a solution of the Bernoulli equation is given by $x(t) = v(t)^{\frac {1}{1-p}} = \left (-1 +4 e^{t^3/3}\right )^3.$

Exercises

In Exercises ?? – ?? decide whether the given differential equation has homogeneous coefficients or is a Bernoulli equation.

$\dps \frac {dx}{dt} = \frac {x^2}{t^3}$ .

$\dps \frac {dx}{dt} = t$ .

$\dps \frac {dx}{dt} = \frac {\cos t}{x^4}\sqrt {t}+x^2$ .

$\dps \frac {dx}{dt} = \frac {t^2}{\sqrt {x}}+x\sin t$ .

$\dps \frac {dx}{dt} = \frac {x}{t} +\frac {\sqrt {t}}{\sqrt {x}}$ .

In Exercises ?? – ?? solve the given initial value problem by an appropriate solution technique.

$\dps \frac {dx}{dt} = \sec \left (\frac {x}{t}\right )+\frac {x}{t}$ where $x(1)=\pi$ .

$\dps \frac {dx}{dt} = x+x^2$ where $x(2)=1$ .

$\dps \frac {dx}{dt} = -\frac {x}{t}-t^3 x^3$ where $x(1)=1$ .

$\dps \frac {dx}{dt} = \frac {x(x+t)}{t^2}$ where $x(1)=1$ .

We set $v(t) = x(t)^{1-p}$ to transform the Bernoulli equation (??) into the equation $\frac {dv}{dt} = (1-p) r(t) v + (1-p) s(t).$ Now set $w(t) = v(\beta t)$ and show that $\beta$ can be chosen so that $w(t)$ is a solution to the linear differential equation $\frac {dw}{dt} = r\left (\frac {t}{1-p}\right ) w + s\left (\frac {t}{1-p}\right ).$