### 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.

*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.

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 \[\frac{\dd y}{\dd x}+{{\cos \left(3\cdot x\right)}\over{\left(x-2\right)\cdot \left(x+9 \right)}}\cdot y=-{{7\cdot x^3}\over{\left(x-2\right)\cdot \left(x+9\right)}}\]In this case we have \[\begin{array}{rcl}\displaystyle p(x)&=&\displaystyle {{\cos \left(3\cdot x\right)}\over{\left(x-2\right)\cdot \left(x+9 \right)}}\\&&\text{ and}\\ \displaystyle q(x)&=&\displaystyle -{{7\cdot x^3}\over{\left(x-2\right)\cdot \left(x+9\right)}}\end{array}\] We note the following regarding the continuity of \(p\) and \(q\).

- The function \(p(x)\) is continuous on the three open intervals \(\left(-\infty,-9\right)\), \(\left(-9,2\right)\) and \(\left(2,\infty\right)\) (i.e., it is undefined at \(x=-9\) and \(x=2\) and continuous elsewhere).

- The function \(q(x)\) is continuous on the three open intervals \(\ivoo{-\infty}{-9}\), \(\ivoo{-9}{2}\), and \(\ivoo{2}{\infty}\) (i.e., it is undefined at \(x=-9\) and \(x=2\) and continuous elsewhere).

All that remains is to determine which of these intervals contains the initial value #x_0# of the initial value problem. Since \(x_0=-4\) and \(-4\in\ivoo{-9}{2}\) it follows that the interval of validity is \(\ivoo{-9}{2}\).

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