Ordinary Differential Equation by substitution.

In summary: That's a "separable" equation.In summary, the conversation discusses solving a given differential equation using a substitution method. The individual attempts to solve the equation by using algebra and making a substitution, but struggles to find a common factor to cancel. Another individual suggests a different approach, integrating the equation with respect to x and y and equating the arbitrary functions. Finally, a third individual suggests using a substitution method involving u and v variables to make the equation separable.
  • #1
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Homework Statement


dy/dx = (x+3y)/(3x+y)

I have to solve the given differential equation by using an appropriate substitution...


The Attempt at a Solution



I used algebra to make the equation (3x+y)dy - (x+3y)dx = 0

then made x = vy and dx = vdy + ydv.

then plugged into get (3vy+y)dy - (vy+3y)(vdy+ydv)

and now I'm looking for a common factor to cancel and I'm stuck, any help? Am I even headed in the right direction...thanks.
 
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  • #2
It doesn't seem you're heading in the right direction. Here's what I would try.

Starting with this equation,
(3x+y)dy = (x+3y)dx

integrate the left side with respect to y. Since you're integrating with respect to y, you'll end up with whatever you get plus an arbitrary function of x alone.
Then, integrate the right side with respect to x, similar to above. This time around you'll need to include an arbitrary function of y alone.

The left and right sides have to be equal for all pairs of (x, y) values, so you can equate the arbitrary functions of x or y on either side with what's on the opposite side.

This should give you the equation h(x, y) = 0.

To check your answer, take the derivative implicitly.
 
  • #3
Looking at the symmetry of "3x+ y" and "x+ 3y", which has "average" value 2x+ 2y= 2(x+ y), let u= x+ y and v= x- y. Then x= (1/2)u+ (1/2)v and y= (1/2)u- (1/2)v, dx= (1/2)du+ (1/2)dv, dy= (1/2)du- (1/2) dv. Put those into the equation. you should wind up with udv= vdu.
 

1. What is an ordinary differential equation?

An ordinary differential equation is a mathematical equation that relates a function or a set of functions to its derivatives. It involves one or more independent variables and one or more dependent variables, and their derivatives with respect to the independent variable.

2. What is substitution in the context of ordinary differential equations?

In the context of ordinary differential equations, substitution refers to the technique of replacing a variable or a set of variables in a differential equation with a new variable or a set of variables, in order to simplify the equation or make it easier to solve.

3. How is substitution used to solve ordinary differential equations?

Substitution is used to solve ordinary differential equations by replacing the original variables with a new set of variables that make the equation easier to solve. This new set of variables is usually chosen based on the structure of the equation and the desired outcome of the solution.

4. What are the advantages of using substitution in solving ordinary differential equations?

Using substitution in solving ordinary differential equations can make the equations simpler and easier to solve. It can also help in obtaining a general solution that can be applied to a wider range of problems. Additionally, substitution can provide more insight into the behavior of the solution and the underlying system.

5. Are there any limitations of using substitution in solving ordinary differential equations?

Although substitution can be a powerful tool in solving ordinary differential equations, it may not always yield a solution. Some equations may not have an explicit solution, and substitution may not be able to simplify them to a solvable form. In such cases, other techniques or numerical methods may need to be employed.

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