what is the differential equation of (1/x-y)dx- (1/y+x)dy=0. Kindly help me. I'm cramming for my exams and there were a lot of problem sets to be done. I hope you guys will help me with these. Thanks a lot in advance.
To find the differential equation of the given expression (1/x-y)dx- (1/y+x)dy=0, we can rearrange the terms to isolate dy/dx.
To do this, let's start by multiplying through by (x^2 + y^2) to simplify the equation:
(x^2 + y^2)(1/x - y)dx - (x^2 + y^2)(1/y + x)dy = 0
Expanding the equation, we have:
x(dx) - xy(dx) - y^2(dx) - x(dy) - y(dy) - xy(dy) = 0
Now, let's group the terms:
(x - xy - y^2)dx + (-x - y - xy)dy = 0
Next, let's rearrange the equation by isolating dy/dx:
(-x - y - xy)dy = -(x - xy - y^2)dx
Divide both sides of the equation by (x - xy - y^2) and multiply by -1:
dy/dx = (x + y + xy)/(x - xy - y^2)
This is now the differential equation for the given expression.
It is important to note that finding the differential equation of an expression is a mathematical process that involves rearranging terms and isolating the derivative. Understanding how to manipulate the equation algebraically and applying the rules of calculus are essential in solving differential equations.