### 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 this case we have \[\begin{array}{rcl}\displaystyle p(x)&=&\displaystyle {{x^2-7\cdot x}\over{x^2+8\cdot x-9}}={{x^2-7\cdot x}\over{\left(x-1\right)\cdot \left(x+9\right)}}\\&&\text{ and}\\ \displaystyle q(x)&=&\displaystyle {{3\cdot \sin \left(3\cdot x\right)}\over{x^2+8\cdot x-9}}={{3\cdot \sin \left(3\cdot x\right)}\over{\left(x-1\right)\cdot \left(x+9\right)}}\end{array}\] We note the following regarding the continuity of \(p\) and \(q\).

- The function \(p(x)\) is continuous on the three open intervals \(\left(-\infty,-9\right)\), \(\left(-9,1\right)\) and \(\left(1,\infty\right)\) (i.e., it is undefined at \(x=-9\) and \(x=1\) and continuous elsewhere).

- The function \(q(x)\) is continuous on the three open intervals \(\ivoo{-\infty}{-9}\), \(\ivoo{-9}{1}\), and \(\ivoo{1}{\infty}\) (i.e., it is undefined at \(x=-9\) and \(x=1\) 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 \(x_0=-2\) and \(-2\in\ivoo{-9}{1}\) it follows that the interval of validity is \(\ivoo{-9}{1}\).

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