Understanding First Order ODEs and Intersection of Curves

  • Thread starter Thread starter ka_reem13
  • Start date Start date
  • Tags Tags
    Ode Stuck
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
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 [itex]\dfrac{dy}{dx}[/itex]. How many real roots does it have?

(b) If two lines [itex]y = m_1x + c_1[/itex] and [itex]y = m_2x + c_2[/itex] intersect, then the angle between them at the intersection is given by [tex]\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}}.[/tex] What is [itex]\cos \theta[/itex] 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:
Reply
  • Like
Likes   Reactions: PeroK
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:
$$\frac{dp}{dx} = \frac{\cancel dp}{\cancel dx}$$
 
Reply
  • Like
Likes   Reactions: erobz, haruspex and SammyS