You are here:

Advanced Math/Differential Equations


what can you say about the existance and uniqness of solution for the below initial value problem without solving it,

first derivative of y(x)=Absolute value of y(x) and y(0)=0.

I didn't know the answer, so I forwarded your question to a math group and got these responses:

The unique solution is y = 0.
I can't do this problem without solving it: it's plain that the unique solution family is y = Ce^x where C >= 0. But y(0) = 0 forces C = 0.

If f(x, y) and ∂f/∂y are both continuous in a neighbourhood of (x₀, y₀), then dy/dx = f(x, y), y(x₀) = y₀ has a unique solution in a neighbohood of (x₀, y₀). This is the fundamental existence theorem for differential equations. Now f(x, y) = |y| is continuous in some open inteval about 0, but ∂f/∂y isn't. So we cannot conclude from the existence theorem that a solution exists, much less that there is a unique solution. The trouble here, however, is that the existence theorem is not "if and only if". It has an "if" part but no "only if". Intuitively I don't think your equation has a solution in any open interval about 0.

Advanced Math

All Answers

Answers by Expert:

Ask Experts


Janet Yang


I can answer questions in Algebra, Basic Math, Calculus, Differential Equations, Geometry, Number Theory, and Word Problems. I would not feel comfortable answering questions in Probability and Statistics or Topology because I have not studied these in depth.


I tutor students (fifth through twelfth grades) and am a Top Contributor on Yahoo!Answers with over 24,000 math solutions.

Co-author of An Outline of Scientific Writing: For Researchers With English as a Foreign Language

I have a Bachelor's degree in Applied Mathematics from the University of California at Berkeley.

Past/Present Clients
George White Elementary School. Homework Help program at the Ridgewood Public Library, Ridgewood, NJ. Individual students.

©2017 All rights reserved.