Understanding First Order ODEs and Intersection of Curves

  • Thread starter Thread starter ka_reem13
  • Start date Start date
  • Tags Tags
    Ode Stuck
Click For Summary
The discussion focuses on solving first-order ordinary differential equations (ODEs) and understanding the intersection of curves. The original poster introduces a perimeter parameter and rearranges the equation to express y in terms of x, leading to a solution involving quadratic curves. However, they struggle to find a second set of curves for the intersections required in part b of the problem. The conversation also touches on calculating the angle between intersecting lines and the gradients of tangent lines at intersection points. Overall, the participants are working through the complexities of ODEs and curve intersections.
ka_reem13
Messages
4
Reaction score
0
Homework Statement
(a) Solve the differential equation:

[x * (dy/dx)^2] - [2y*(dy/dx)] - x = 0

How many integral curves pass through each point of the (x,y) plane (except x = 0)?
why is the solution at each point not unique

(b) The differential equation:
[(dy/dx)^2] + [f(x,y)*(dy/dx)] - 1 = 0
represents a set of curves such that two curves pass through any given point. Show that these curves intersect at right angles at the point. at f = -2y/x verify this property for the point (3,4)
Relevant Equations
differential equations
I'm aware that I can introduce the perimeter p = dy/dx
then I can rearrange my equation to make y the subject, then I can show that dp/dx = p/x. However, this only gives me a bunch of quadratic curves for my solution. However given part b I see that two curves are meant to intersect each point and I don't know where I'll get the second set of curves (solutions) from.

for part b honestly I don't even know where to start
 
Physics news on Phys.org
(a) The ODE is a quadratic in \dfrac{dy}{dx}. How many real roots does it have?

(b) If two lines y = m_1x + c_1 and y = m_2x + c_2 intersect, then the angle between them at the intersection is given by \cos \theta = \frac{(1,m_1)\cdot(1,m_2)}{\|(1,m_1)\|\|(1_,m_2)\|} = \frac{1 + m_1m_2}{\sqrt{1 + m_1^2}\sqrt{1 + m_2^2}}. What is \cos \theta if the lines intersect at right angles? To apply this to two curves, one looks at the tangent lines at the point of intersection. What are the gradients of these tangent lines if the curves are the integral curves of this ODE?
 
Last edited:
ka_reem13 said:
introduce the perimeter
parameter
 
ka_reem13 said:
I can show that dp/dx = p/x
You can? I don’t see how. Please post your working.
 
ka_reem13 said:
I can show that dp/dx = p/x
haruspex said:
You can? I don’t see how.
Easy, just cancel the d's. :oldbiggrin:
$$\frac{dp}{dx} = \frac{\cancel dp}{\cancel dx}$$
 
  • Like
Likes erobz, haruspex and SammyS
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...