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.)

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 must satisfy the initial condition . This means that the solution y(x) must pass through the point on the plane .

In this case, we know that is already a solution of the DE xy'+y=2x, the value of and we have to find the values for which the equality holds, that is,

Multiplying by the above equation we obtain , where it follows by susbtracting that This is a second-degree polynomial equation and its solutions can be found by factoring: . Therefore, there are two possible values for : 7 and -1. From here we can see that -1 is the smaller and 7 is the larger.

Now, the larger interval for which is a solution of the IVP (the smaller value of ) is the maximal interval where y(x) is defined and pass through the point (-1,6). In this case, it must be that .

Finally, the larger interval for which is a solution of the IVP (the larger value of ) is the maximal interval where y(x) is defined and pass through the point (7,6). In this case, it must be that .

where's the diagram?

step-by-step explanation: