Question

In Problems 9 and 10 find an interval centered about $\mathbf{x}\mathbf{=}\mathbf{0}$for which the given initial-value problem has a unique solution.

$\left(x-2\right)\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The interval for the unique solution is $\left(-\infty ,2\right)$.

Step-by-step solution

Define general second-order linear differential equation and the Theorem for Existence of a Unique Solution

A general second-order linear differential equation is defined as ${\mathrm{a}}_2\left(\mathrm{x}\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\mathrm{a}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{a}}_0\left(\mathrm{x}\right)\mathrm{y}=\mathrm{g}\left(\mathrm{x}\right)$.

Consider${\mathrm{a}}_{\mathrm{n}}\left(\mathrm{x}\right),{\mathrm{a}}_{\mathrm{n}-1}\left(\mathrm{x}\right),....,{\mathrm{a}}_1\left(\mathrm{x}\right),{\mathrm{a}}_0\left(\mathrm{x}\right)$ and$\mathrm{g}\left(\mathrm{x}\right)$ be continuous on an interval I and let${\mathrm{a}}_{\mathrm{n}}\left(\mathrm{x}\right)\ne 0$ for every value of x in this interval. If there is any point$\mathrm{x}={\mathrm{x}}_0$ in this interval, then there exists a solution$\mathrm{y}\left(\mathrm{x}\right)$ of this initial-value problem on the interval and is unique. This is known as the Theorem for Existence of a Unique Solution.

Find the interval in which the functions are continuous

Comparing the given differential equation with the general second-order linear differential equation,

$\left(\mathrm{x}-2\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+3\mathrm{y}=\mathrm{x}$

The functions are:

$$\begin{gathered} {\mathrm{a}}_2\left(\mathrm{x}\right)=\mathrm{x}-2 \\ {\mathrm{a}}_1\left(\mathrm{x}\right)=0 \\ {\mathrm{a}}_0\left(\mathrm{x}\right)=3 \\ \mathrm{g}\left(\mathrm{x}\right)=\mathrm{x} \end{gathered}$$

The last three functions are continuous in the interval$\left(-\infty ,\infty \right)$and the first function is continuous in the intervals$\left(-\infty ,2\right)$and $\left(2,\infty \right)$.

Out of these intervals,$\left(-\infty ,2\right)$contains $\mathrm{x}=0$.

Find the interval for unique solution

All the functions are continuous in the interval for ${\mathrm{a}}_2\left(\mathrm{x}\right)\ne 0$,true for all values of $\mathrm{x}$.

Using the Theorem for Existence of a Unique Solution,

localid="1668489706979" $\mathbf{Hence}\mathbf{,}\mathbf{}\mathbf{The}\mathbf{}\mathbf{interval}\mathbf{}\mathbf{for}\mathbf{}\mathbf{the}\mathbf{}\mathbf{initial}\mathbf{-}\mathbf{value}\mathbf{}\mathbf{problem}\mathbf{}\mathbf{having}\mathbf{}\mathbf{a}\mathbf{}\mathbf{unique}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{is}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{.}$