Uniqueness of solutions of linear first-order ODEs

Differential equations: Linear first-order differential equations

Uniqueness of solutions of linear first-order ODEs

As we know, linear first-order differential equations have the following form, where $a(t)$ , $b(t)$ , and $f(t)$ are functions with $a(t)\ne0$ : $a(t)\cdot \frac{\dd y}{\dd t}+b(t)\cdot y=f(t)$ It is also known that the equation is called homogeneous if $f(t)=0$ . The functions $a(t)$ and $b(t)$ are called coefficients and $f(t)$ the inhomogenous term. If these coefficients are constant, the general solution can be expressed in terms of standard functions.

We may assume that the coefficient $a(t)$ is distinct from zero (that is, the constant function $0$ ). For otherwise the differential equation would be of first order. Therefore, we can divide by $a(t)$ . In the resulting equation, the coefficient of $y'$ is equal to $1$ . In this case, we say that the equation is in standard form.

Uniqueness of solutions of linear first-order differential equations

Let $t_0$ be a point in an open interval $\ivoo{c}{d}$ (that is to say: $c\lt t_0\lt d$ ) and let $p$ and $q$ be continuous functions on this interval. Then the initial value problem $y' + p(t)\cdot y= q(t), \phantom{xxx}\phantom{xx}y(t_0) = \alpha$ where $\alpha$ is an arbitrary number, has a unique solution defined on the entire interval $\ivoo{c}{d}$ .

The extremes $c$ and $d$ may be equal to $-\infty$ and $\infty$ , respectively.

A consequence of the theorem is the fact that the general solution of a linear first-order ODE with the specified properties has a single free parameter (integration constant). In terms of linear algebra: the solution set is $1$ -dimensional.

A very simple example is $y'=q(t)$ , where $q$ is a continuous function defined on an open interval $0$ . A particular solution is found by calculating an anti-derivative of the right hand side (it exists because $b$ is continuous). The corresponding homogeneous equation has the general solution $y=C$ . This can be seen directly by taking an anti-derivative. But we can also deduce this fact from the theorem: Let $y$ be any solution of the homogeneous ODE $y'=0$ . Choose $\alpha=y(0)$ . The theorem shows that there is exactly one function $u$ with $u'=0$ and $u(0)=\alpha$ . Both $y$ and the constant function $\alpha$ meet these conditions. This forces $y=u=\alpha$ . It shows that every solution of the homogeneous ODE has the form $C$ . The conclusion is that, if $Q(t)$ is an anti-derivative of $q(t)$ , the general solution of the original ODE , $y'=q(t)$ , is:

$y= Q(t)+C$

This solution is indeed uniquely determined if we specify $y(0)=\alpha$ : then we have $y(t) = Q(t)+\alpha-Q(0)$ . This solution is defined on the entire interval $\ivoo{c}{d}$ .

The proof is outside the scope of this course. In the special case $y'=q(t)$ the theorem follows from known facts on integration.

Consider the initial value problem $(x-6)\cdot\frac{\dd y}{\dd x}+5\cdot \e^{2\cdot x}\cdot y=9\cdot x^2,\qquad y(0)=-15$ Use Uniqueness of the solution of a linear first-order ODE to determine the interval of validity of the initial value problem.

Enter the interval in the answer field by using the interval buttons under the function tab of the input pallette. The symbol $\infty$ can also be found under this tab of the pallette.

Interval of validity = $\ivoo{-\infty}{6}$

The standard form of a linear first-order initial value problem is $\frac{\dd y}{\dd x}+p(x)\cdot y=q(x),\qquad y(x_0)=\alpha$ The Uniqueness of the solution of a linear first-order ODE says that the solution to the above initial value problem exists and is unique on the largest open interval $\ivoo{c}{d}$ containing $x_0$ on which both $p(x)$ and $q(x)$ are continuous. This is the interval of validity.

In order to use the existence and uniqueness theorem we first rewrite the differential equation in the standard linear form $\frac{\dd y}{\dd x}+{{5\cdot \e^{2\cdot x}}\over{x-6}}\cdot y={{9\cdot x^2}\over{x-6}}$ In this case we have $\begin{array}{rcl}\displaystyle p(x)&=&\displaystyle {{5\cdot \e^{2\cdot x}}\over{x-6}}\\&&\text{ and}\\ \displaystyle q(x)&=&\displaystyle {{9\cdot x^2}\over{x-6}}\end{array}$ We note the following regarding the continuity of $p$ and $q$ .

The function $p(x)$ is continuous on the two open intervals $\left(-\infty,6\right)$ and $\left(6,\infty\right)$ (i.e., it is undefined at $x=6$ and continuous elsewhere).

The function $q(x)$ is continuous on the two open intervals $\ivoo{-\infty}{6}$ and $\ivoo{6}{\infty}$ (i.e., it is undefined at $x=6$ and continuous elsewhere).

Hence $p(x)$ and $q(x)$ are simultaneously continuous on the two open intervals $\ivoo{-\infty}{6}$ and $\ivoo{6}{\infty}$ .

All that remains is to determine which of these intervals contains the initial value $x_0$ of the initial value problem. Since $x_0=0$ and $0\in\ivoo{-\infty}{6}$ it follows that the interval of validity is $\ivoo{-\infty}{6}$ .

New example