Also, I am aware of the fact that the equation is a separable 1st order ordinary differential equation, but by substituting the factor e^tan-1(x), the answer will be in "e" format, which has no relevance to the given answers. There has to be a trick, or maybe I am just confusing myself even more.
All that I know, is that I have to find dy/dx from the equation x^2+1=y/(x-dy/dx).
The multiple choice answers are the following:
a)dy/dx= 2(x+y)^2+ (x/y)
e)none of the above
I have tried the advice hunt_mat has...
I think that I am supposed to calculate dy/dx by thinking that the equation given, is separable equation or linear. This is a first order differential equation, thus the answer cannot be this simple...
Thanks for your contribution though.
If x^2+1=y/(x-y'), where y'=dy/dx, find dy/dx
I have tried so many ways, but I cannot seem to get the correct answer.
The answers I have got previously are:
x² + 1 = y/(x - y')
(x² + 1)(x - y') = y
x(x² + 1) - y'(x² + 1) = y
x(x² + 1) - y = y'(x² + 1)...
Also, i) the answers are that left limit is 1, and the right limit is 2, and that g(x) is continuous at x=1;
ii) that left limit is 2, and the right limit is 1, and that g(x) is not differentiable at x=1;
iii) that left limit is 1, and the right limit is 2, and that g(x) is not differentiable...
However, I am still unsure what the limits would be. I reckon that both left limit (x^2 + 2 if x<=1) and right limit( x + 2 if x>1), should be infinite.
Correct me if I am wrong.
Though, at x=1, the limit would be 3 right?
g(x) = [ x^2 + 2 if x<=1 & x + 2 if x>1,
I am asked to find the left and right limits, and whether the g(x) is continuous or not.
The Attempt at a Solution
When I draw the two equations, I get a hyperbola and a line of gradient 1. They...