Given that y = \frac {1}{2-x} is a solution of the IVP y' - y^2 = 0, y(1) = 1 Find the largest...


Given that {eq}y = \frac {1}{2-x} {/eq} is a solution of the IVP

{eq}y' - y^2 = 0 {/eq}, {eq}y(1) = 1 {/eq}

Find the largest interval of the solution.

Finding Interval of Solution

The question provides us a solution of a first order ordinary differential equation (ODE) with an initial condition (IC) that comprises an initial value problem (IVP). Looking at the solution and the IC we determine the largest interval where a solution of the IVP exists.

Answer and Explanation: 1

Become a member to unlock this answer!

View this answer

Since the solution is given as

{eq}\displaystyle y=\frac {1}{2-x} \qquad (1) {/eq}

we note from (1) that owing to division by zero, {eq}x \ne...

See full answer below.

Learn more about this topic:

Existence Proofs in Math: Definition & Examples


Chapter 3 / Lesson 4

Examine existence theorems in mathematics. Learn how to construct an existence proof. Study some of the most important existence theorems and see examples.

Related to this Question

Explore our homework questions and answers library