How can I solve this second order differential equation?

  • Thread starter raul_l
  • Start date
In summary, "Dx/dt" represents the rate of change of the variable x with respect to time, while "dy/dt" represents the rate of change of the variable y with respect to time. The value "1-1/y" represents the instantaneous rate of change of y with respect to x, and the equation is nonlinear and can be solved using separation of variables. This type of differential equation has various applications in physics, engineering, and economics.
  • #1
raul_l
105
0

Homework Statement



dx/dt=1-1/y
dy/dt=1/(x-t)

The Attempt at a Solution



If I take the derivative of the second equation and substitute it to the first one I

[tex] \frac{d^2 y}{dt^2} - \frac{1}{y} (\frac{dy}{dt})^2 = 0 [/tex]

but I don't know how to solve this. Could anyone name any methods that could try?
Thanks.
 
Physics news on Phys.org
  • #2
The trick is to notice
[tex]
y''-\frac{1}{y}(y')^{2}=0\Rightarrow\frac{y''}{y'}=\frac{y'}{y}
[/tex]
 
  • #3
[tex]
y''-\frac{1}{y}(y')^{2}=0\Rightarrow yy''= y'^{2} [/tex]

Now put in the equation [itex] y=Ae^{Bt} [/itex] and find A and B.
 

1. What does "Dx/dt" and "dy/dt" represent in this equation?

"Dx/dt" represents the rate of change of the variable x with respect to time, while "dy/dt" represents the rate of change of the variable y with respect to time. Essentially, they represent the derivatives of x and y, respectively, with respect to time.

2. What does the value "1-1/y" mean in this equation?

The value "1-1/y" represents the instantaneous rate of change of y with respect to x. This means that for every unit increase in x, there is a corresponding decrease in the rate of change of y. This is also known as the inverse relationship between x and y.

3. Is this equation a linear or nonlinear differential equation?

This equation is nonlinear, as it contains variables with powers other than 1 and does not follow the form of a linear differential equation (where the dependent variable and its derivatives are only raised to the first power).

4. How can this equation be solved?

To solve this equation, we can use the method of separation of variables. This involves isolating the variables on opposite sides of the equation and then integrating both sides to find the solution.

5. What are the applications of this differential equation?

This type of differential equation is commonly used in fields such as physics, engineering, and economics to model and analyze various real-world phenomena. It can be used to study rates of change, growth, and decay in systems that exhibit an inverse relationship between two variables.

Similar threads

  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
12
Views
991
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
840
  • Calculus and Beyond Homework Help
Replies
19
Views
776
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
130
Back
Top