# Stuck on a 1st order ODE

• ka_reem13
In summary, the conversation discusses introducing the perimeter as a parameter and rearranging the equation to make y the subject. However, this only results in a bunch of quadratic curves for the solution. Part b mentions finding the second set of curves for the solution and using the tangent lines at the point of intersection to determine the gradients of the integral curves. The conversation ends with a discussion on how to show that dp/dx = p/x.

#### ka_reem13

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

(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:
PeroK
ka_reem13 said:
introduce the perimeter
parameter

ka_reem13 said:
I can show that dp/dx = p/x
$$\frac{dp}{dx} = \frac{\cancel dp}{\cancel dx}$$