- #1
aatkins09
- 7
- 0
dy/dx=(-x+sqrt(x^2+y^2))/y
USE SUBSTITUTION u=x^2+y^2
I tried to work it out but got a really ugly answer, please help!
USE SUBSTITUTION u=x^2+y^2
I tried to work it out but got a really ugly answer, please help!
Nex Vortex said:If you use the correct substitution then several things should cancel. It would be helpful to also post your attempt at the solution in order to see where you are having problems.
uart said:Instead of starting out by trying to separate and integrate, try finding an expression for du/dx and using it to replace dy/dx
aatkins09 said:du/dx would be 2x+2y but I don't understand where to go from there. how can I replace dy/dx with it?
Nex Vortex said:Remember that you have to use implicit differentiation.
[tex]du=2xdx+2ydy[/tex]
Then you can solve for [tex]dy/dx[/tex]
You should find that when you find this and substitute, several things should cancel.
Nex Vortex said:Remember that you have to use implicit differentiation.
[tex]du=2xdx+2ydy[/tex]
Then you can solve for [tex]dy/dx[/tex]
You should find that when you find this and substitute, several things should cancel.
aatkins09 said:nvm, I still can't get it. it's okay though, thank you, I appreciate all of your help!
uart said:[tex] \frac{du}{dx} = 2x+2y \, \frac{dy}{dx}[/tex]
Now just substitute in your original expression for dy/dx and it literally just falls into place.
The notation "dy/dx" represents the derivative of the function y with respect to x. It measures the rate of change of y with respect to x, or how much y changes for a small change in x.
This is a differential equation, which means it involves derivatives. To solve it, you can use various methods such as separation of variables, substitution, or integrating factors. The specific method will depend on the form of the equation and your personal preference.
The variable "u" represents the independent variable in this equation. It may be a function of x or y, but it is not explicitly stated. This allows us to solve for the general solution of the equation, rather than just a specific solution for a given value of u.
Yes, there may be multiple solutions to this equation. In general, a differential equation has an infinite number of solutions. However, the specific solution for a given initial condition can be determined by specifying the value of y when x is known.
Yes, this equation can be used to model real-world situations where the rate of change of a quantity depends on the value of that quantity and other variables. For example, it can be used to model population growth, chemical reactions, and other physical processes.