Question

The function y = x − 7 x is a solution of the DE xy' + y = 2x. Find x0, given the first-order IVP xy' + y = 2x, y(x0) = 6. Enter your answers as a comma-separated list. x0 = Find the largest interval I for which y(x) is a solution of the IVP for the smaller value of x0. (Enter your answer using interval notation.) Find the largest interval I for which y(x) is a solution of the IVP for the larger value of x0. (Enter your answer using interval notation.)

Answer 1

`x_0=\{-1,7\}, I_s=(-\infty,0), I_l=(0,+\infty)`

Explanation:

Remember that an IVP (initial value problem) of first order consists of solving a first-order differential equation (DE) and the solution
`y(x)` must satisfy the initial condition
`y(x_0)=y_0`. This means that the solution y(x) must pass through the point on the plane
`(x_0,y_0)` .

In this case, we know that
`y(x)=x-(7)/(x)` is already a solution of the DE xy'+y=2x, the value of
`y_0=6` and we have to find the values
`x_0` for which the equality
`y(x_0)=6` holds, that is,
`y(x_0)=x_0-(7)/(x_0)=6.`

Multiplying by
`x_0` the above equation we obtain
`6x_0=x_0^2-7`, where it follows by susbtracting
`6x_0` that
`\\x_0^2-6x_0-7=0.` This is a second-degree polynomial equation and its solutions can be found by factoring:
`x_0^2-6x_0-7=(x_0-7)(x_0+1)=0`. Therefore, there are two possible values for
`x_0`: 7 and -1. From here we can see that -1 is the smaller and 7 is the larger.

Now, the larger interval
`I_s` for which
`y(x)` is a solution of the IVP
`xy'+y=2x, y(-1)=6` (the smaller value of
`x_0`) is the maximal interval where y(x) is defined and pass through the point (-1,6). In this case, it must be that
`I_s=(-\infty, 0)`.

Finally, the larger interval
`I_l` for which
`y(x)` is a solution of the IVP
`xy'+y=2x, y(7)=6` (the larger value of
`x_0`) is the maximal interval where y(x) is defined and pass through the point (7,6). In this case, it must be that
`I_l=(0,+\infty)`.