Answer:
[tex]x_0=\{-1,7\}, I_s=(-\infty,0), I_l=(0,+\infty)[/tex]
Step-by-step explanation:
Remember that an IVP (initial value problem) of first order consists of solving a first-order differential equation (DE) and the solution [tex]y(x)[/tex] must satisfy the initial condition [tex]y(x_0)=y_0[/tex]. This means that the solution y(x) must pass through the point on the plane [tex](x_0,y_0)[/tex] .
In this case, we know that [tex]y(x)=x-\dfrac{7}{x}[/tex] is already a solution of the DE xy'+y=2x, the value of [tex]y_0=6[/tex] and we have to find the values [tex]x_0[/tex] for which the equality [tex]y(x_0)=6[/tex] holds, that is, [tex]y(x_0)=x_0-\dfrac{7}{x_0}=6.[/tex]
Multiplying by [tex]x_0[/tex] the above equation we obtain [tex]6x_0=x_0^2-7[/tex], where it follows by susbtracting [tex]6x_0[/tex] that [tex]\\x_0^2-6x_0-7=0.[/tex] This is a second-degree polynomial equation and its solutions can be found by factoring: [tex]x_0^2-6x_0-7=(x_0-7)(x_0+1)=0[/tex]. Therefore, there are two possible values for [tex]x_0[/tex]: 7 and -1. From here we can see that -1 is the smaller and 7 is the larger.
Now, the larger interval [tex]I_s[/tex] for which [tex]y(x)[/tex] is a solution of the IVP [tex]xy'+y=2x, y(-1)=6[/tex] (the smaller value of [tex]x_0[/tex]) is the maximal interval where y(x) is defined and pass through the point (-1,6). In this case, it must be that [tex]I_s=(-\infty, 0)[/tex].
Finally, the larger interval [tex]I_l[/tex] for which [tex]y(x)[/tex] is a solution of the IVP [tex]xy'+y=2x, y(7)=6[/tex] (the larger value of [tex]x_0[/tex]) is the maximal interval where y(x) is defined and pass through the point (7,6). In this case, it must be that [tex]I_l=(0,+\infty)[/tex].
