Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I had a question about how to do one of these problems. So here's the question:
Given this equation $y'=\frac{-\cos(t)y(t)}{(t+2)(t-1)}+t$, find if the initial conditions $y(0)=10, y(2)=-1, y(-10)=5$ exist.

So I think the first step is just to take the partial derivative with respect to y which gives me: $$y''=\frac{-\cos(t)y'(t)}{(t+2)(t-1)}$$

So the 1'st equation doesn't exist at $t=-2,1$ and the partial derivative doesn't exist at $t=-2,1$ ....so do I conclude that all the initial values exists since none of them are $y(-2)$ or $y(1)$.

Don't really know how to do this whole existence and uniqueness thing....so am I right or completely off track?

share|improve this question
The initial conditions exist, because you have written them down. The question is whether there exist unique solutions, given the DE and the initial conditions. –  Christian Blatter Feb 7 '13 at 11:46
add comment

1 Answer

up vote 2 down vote accepted

What you've done looks perfectly fine.

Here's a general outline of what you do when you're looking to find where solutions exist for the first-order differential equation $y'+ p(t) y = g(t)$.

  1. Write the differential equation in the form: $y'=f(y, t)$

  2. Find $f_y = \frac{\partial}{\partial y} f$.

  3. Determine points of discontinuities of both $f_y$ and $f$.

At this point, if you're just looking to see if a particular initial condition ($t_0$) has a solution, just check if $t_0$ is one of the points of discontinuity.

If you're looking for where the solution exists:

  1. Draw a number line denoting where the discontinuities are (if possible).
  2. Find where the initial condition falls on the number line.
  3. If the discontinuity to the left of $t_0$ is $a$, and the discontinuity to the right of $t_0$ is $b$, then the solution exists on the interval $(a, b)$.

EDIT Based on requests from comments below, here's a statement of the existence and uniqueness theorem:

Let the functions $f$ and $\frac{\partial f}{\partial y}$ be continuous in some rectangle $\alpha < t < \beta$, $\gamma < y < \delta$ containing the point $(t_0, y_0)$. Then, in some interval $t_0 - h < t < t_0 + h$ contained in $\alpha < t < \beta$, there is a unique solution $y = \phi(t)$ of the initial value problem:

$$\begin{array}{cc} y' = f(t,y) & y(t_0) = y_0. \end{array}$$

Source: Elementary Differential Equations and Boundary Value Problems, Boyce and DiPrima, 10th Edition, pg 70.

share|improve this answer
what happens if $f$ is discontinuous at $y=2$ but $f_y$ is discontinuous at $y=3$? What will I say about the existence of $y(2)$ and $y(3)$? –  Charlie Yabben Feb 7 '13 at 3:00
@CharlieYabben The uniqueness and existence theorem assumes $f$ and $f_y$ are continuous on an interval containing $t_0$. Thus, in your example from the previous comment, conclude that $y(2)$ and $y(3)$ do not exist. –  anorton Feb 7 '13 at 3:10
@anorton: It is a good idea to include the statement of the existence and uniqueness theorem or add a link to it. –  Mhenni Benghorbal Feb 7 '13 at 9:05
@MhenniBenghorbal I have now included one. –  anorton Feb 7 '13 at 12:16
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.