Understanding First Order ODEs and Intersection of Curves

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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
 
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(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?
 
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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:
$$\frac{dp}{dx} = \frac{\cancel dp}{\cancel dx}$$
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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