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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?

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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
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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.

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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
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