Solving a Non-Homogeneous Equation: y'=[a/(x+y)]^2

[lewis198]

I've been going through problems in a classic maths for physics book, and am stuck on a question.

I can't find the solution for this equation:

y'=[a/(x+y)]^2

I tried directly integrating it, using iterative techniques, and I know it is not homogeneous so I can't use substitution.

 

[Dick]

It doesn't have to be homogeneous for substitution to be useful. Try u=x+y.

 

[lewis198]

thanks for the reply. What is the criteria for a substitution?

 

[Dick]

thanks for the reply. What is the criteria for a substitution?

The only criterion for a substitution is that it simplifies the equation in such a way that you can solve it. The equation in terms of u is separable.

 

[HallsofIvy]

thanks for the reply. What is the criteria for a substitution?

Whatever works!

 

[lewis198]

Thanks again. I used algebraic long division and trigonometric substitution and got the following solution:

y-arctan((x+y)/a)=C

This is not a so called 'closed' solution. Is that ok? Is this solution even meaningful?

 

[HallsofIvy]

Yes, that's a perfectly good solution. In general, since dx/dy= 1/(dy/dx), for first order differential equations, the distinction between "independent variable" and "dependent variable" doesn't exist and it is seldom possible to solve for one variable "in terms of" the other.

I remember in my first "differential equations" course, early in the course, I arrived at a solution that involved an integral I just could not do. After a long time of trying everything I could think of, I finally looked in the back of the book. I found that the answer was given in terms of that integral! When you working with differential equations, don't expect everything to be simple!

 

[Dick]

Thanks again. I used algebraic long division and trigonometric substitution and got the following solution:

y-arctan((x+y)/a)=C

This is not a so called 'closed' solution. Is that ok? Is this solution even meaningful?

You've got the right technique. I think you might be missing a factor of 'a' though.