Question

The function$\mathit{y}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{/}\mathit{x}$is a solution of the DE$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$. Find${\mathit{x}}_{\mathbf{0}}$and the largest interval Ifor which y(x)is a solution of the first-order IVP$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$,$\mathit{y}\mathbf{\left(}{\mathit{x}}_{\mathbf{0}}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

Short answer

The largest interval of definition over which y(x) is defined, differentiable with continuous derivative and contains the initial point for ${\mathrm{x}}_0=2$is $(0,\infty )$, and for ${\mathrm{x}}_0=-1$is $(-\infty ,0)$.

Step-by-step solution

Define the derivation of a function and an explicit function

An explicit function is defined as one that is expressed in terms of the independent variable.

The sensitivity of the function is derived by the derivative of a function of a real variable for modifying the argument.

Determine the largest interval

The domain of the function $\mathrm{y}=\mathrm{x}-2/\mathrm{x}$is the set of all real numbers x except $x=0$, i.e. $\mathrm{R}-\left\{0\right\}$.

Substitute $\mathrm{y}({\mathrm{x}}_0)=1$in the given function.

$\begin{array}{c}y\left(x_0\right)=x_0-\frac{2}{x_0}\\ 1=x_0-\frac{2}{x_0}\\ x_0^2-x_0-2=0\end{array}$

Solve the above equation by using the quadratic formula.

$\begin{array}{c}{\mathrm{x}}_0=\frac{1\pm \sqrt{1-4\left(1\right)(-2)}}{2}\\ =\frac{1\pm \sqrt{9}}{2}\\ =\frac{1\pm 3}{2}\\ {\mathrm{x}}_0=2\mathrm{or}-1\end{array}$

Hence, the largest interval of definition over which y(x) is defined, and differentiable with continuous derivative and contains the initial point for ${\mathrm{x}}_0=2$is $(0,\infty )$, and for ${\mathrm{x}}_0=-1$is $(-\infty ,0)$.