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I believe this is not a duplicate because the $D$ is different.


The problem: Does

$$x = \frac{1}{4-t^2} \tag{*}$$

yield a solution/s to

$$\frac{dx}{dt} = 2tx^2 \tag{**}$$

If so, give all maximal solutions together with their intervals of definition.


Definition: Let $x = x(t)$. A solution of the first order ODE $x' = f(t,x)$, where $f$ is defined on some domain $D \subseteq \mathbb R^2$ s.t. $D$ is open and connected, is a differentiable function $\varphi$ on some interval $I \subseteq \mathbb R$ s.t.

  1. $(t, \varphi(t)) \in D \ \forall t \in I$

  2. $\varphi'(t) = f(t, \varphi(t)) \ \forall t \in I$


Existence Theorem: If $f$ is continuous on a domain $D \subseteq \mathbb R^2$ and $(\tau, \xi) \in D$, then $\exists$ a solution $\varphi$ of $x' = f(t,x)$ defined on some interval $I$ s.t. $\tau \in I$ and $\varphi(\tau) = \xi$


What I tried:

1. On the choice of D

The denominator vanishes when $t = 2$ or $-2$ so I guess we can't say $D = \mathbb R^2$. How about

$$D = \{(t,x) | t > 2 \}$$

or

$$D = \{(t,x) | 2 > t > -2 \}$$

or

$$D = \{(t,x) | t < -2 \}$$

Are any open and connected? We didn't have any topology courses, and our real analysis only covered $\mathbb R$.


2. On existence

$f = 2tx^2$ is continuous on $D = \{(t,x) | t < -2 \}$ (I'm assuming this is right)

Choose $\xi = 10$ and $\tau = \tau^{10}$ where $x(\tau^{10}) = 10$.

Then $\exists$ a solution $\varphi$ of $x' = f = 2tx^2$ defined on some interval $I$ s.t. $\tau = \tau^{10} \in I$ and $\varphi(\tau^{10}) = 10$


3. On the choice of I

I differentiated $(*)$ to get $(**)$.

The denominator vanishes when $t = 2$ or $-2$ so I think the solutions are:

$$\varphi_1(t) = \frac{1}{4-t^2} \ on \ I = (2, \infty)$$

and

$$\varphi_2(t) = \frac{1}{4-t^2} \ on \ I = (-2, 2)$$

and

$$\varphi_3(t) = \frac{1}{4-t^2} \ on \ I = (-\infty, -2)$$


Is that right?


For another problem

$$x = \frac{2}{1-e^t}, \frac{dx}{dt} = \frac{x(x-2)}{2}$$

do we similarly have

$$\varphi_1(t) = \frac{2}{1-e^t} \ on \ I = (0, \infty)$$

and

$$\varphi_2(t) = \frac{2}{1-e^t} \ on \ I = (-\infty, 0)$$

For what $D$? $D = \{(t,x) | t < 0 \}$ okay? How about $D = \{(t,x) | t > 0 \}$?

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marked as duplicate by Alex M., user296602, Winther, Yagna Patel, Did Jan 30 '16 at 7:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ the general solution is given by $$x(t)=\frac{1}{C-t^2}$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 25 '16 at 17:36
  • $\begingroup$ @Dr.SonnhardGraubner The question is if the $x$ given above yields solutions for the ODE. this question is mainly about existence of solutions and not really solving DEs $\endgroup$ – BCLC Jan 25 '16 at 17:50
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Note that in problems like this you should never use the existence theorem. Simply substitute the function in your equation.

More precisely, clearly $D=\mathbb R^2$ and so you need to check two things: is the function $1/(4-t^2)$ differentiable on some open interval? does it satisfy the equation?

For the first question you can take any open interval contained in $\mathbb R\setminus\{-2,2\}$. For the second question you just substitute in the equation and check that indeed it is a solution.

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