Differential equation first, first degree help

In summary: The partial derivatives will tell you if the equation is exact or not. You can try integrating both sides to see if the equation is exact or not. If it is not exact, then you can try to find a different equation that is exact.
  • #1
elphin
18
0
Differential equation first, first degree help!

Homework Statement



solve: x.dy + y.dx + (x.dy - y.dx)/(x^2 + y^2) = 0

Homework Equations



if M.dx + N.dy = 0 has to be exact then

equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

The Attempt at a Solution



the idea is to find if this equation is exact because once you do that the integration is easy..

but

now i first i simplify the equation by multiplying throughout by (x^2 + y^2)

simplified form: (x^3 + x.y^2 - y).dx + (y^3 + y.x^2 + x).dy = 0
try to check if it satisfies equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

here M = (x^3 + x.y^2 - y) & N = (y^3 + y.x^2 + x)

we get 2.x.y - 1 is not equal to 2.x.y + 1

now i try another method i.e. group the terms without multiplying throughout by (x^2 + y^2)

simplified equation: [x - (y/(x^2 + y^2))].dx + [y + (x/(x^2 + y^2))].dy = 0
now when we apply equation 1 criteria we get:

partial derivative of M w.r.t y (keeping x constant) = (y^2 - x^2)/(x^2 + y^2)^2 = partial derivative of N w.r.t x (keeping y constant) = (y^2 - x^2)/(x^2 + y^2)^2

my question is - why is it that i am getting the two methods to be different?
 
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  • #2


eh... anyone? wold really help.. this has been bugging me...
 
  • #3


Have you seen integrating factors yet? That whole method is based on the idea of multiplying the equation by a function to put it in a 'nice' form. When you have an equation M(x,y)dx + N(x,y)dy=0, multiplying by anything is going to change the derivatives. It's very likely that it will change the partial derivatives of M and N in different ways, which may or may not help you make the equation into an exact equation. For example, if I multiply the equation by x, that's going to change the x partial derivative of N, but it's not going to change the y partial of M.
 
  • #4


elphin said:

Homework Statement



solve: x.dy + y.dx + (x.dy - y.dx)/(x^2 + y^2) = 0

Homework Equations



if M.dx + N.dy = 0 has to be exact then

equation 1: partial derivative of M w.r.t y (keeping x constant) = partial derivative of N w.r.t x (keeping y constant)

The Attempt at a Solution



the idea is to find if this equation is exact because once you do that the integration is easy ...
Notice that, d(xy) = x.dy + y.dx .

Also, d(y/x) = (xdy - ydx)/x2

So, you can rewrite your equation as: (x2 + y2)d(xy) + x2(d(y/x)) = 0

This suggests that you let u = xy and v = x/y.

Can you take it from here?
 

FAQ: Differential equation first, first degree help

What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It is commonly used to describe the rate of change of a system over time.

What is the first degree in a differential equation?

The first degree in a differential equation refers to the highest power of the derivative present in the equation.

Why is it important to solve first degree differential equations?

First degree differential equations are important because they can model many real-world problems such as growth rates, decay rates, and motion of objects. Solving these equations helps us understand and predict the behavior of these systems.

What methods are used to solve first degree differential equations?

There are several methods for solving first degree differential equations, such as separation of variables, integrating factors, and Euler's method. The choice of method depends on the specific equation and initial conditions.

Can differential equations be solved analytically or numerically?

Yes, differential equations can be solved both analytically and numerically. Analytical solutions involve finding an explicit formula for the solution, while numerical solutions involve approximating the solution using numerical methods such as Euler's method or Runge-Kutta methods.

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