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Valentin Fadeev

Going beyond dx

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Edited by Valentin Fadeev, Sunday, 18 Sep 2011, 23:23

This is quite a minor trick and like many things listed here may seem quite trivial. However, this is one of those few occasions when I had the tool in mind, before I actually got the example touse it on. Consider:

equation left hand side d times y divided by d times x equals right hand side one plus two divided by x plus y

which does not really require a great effort to solve. But forget all the standard ways for a moment and add d times x divided by d times x equals one to both parts:

equation left hand side d times open x plus y close divided by d times x equals right hand side two times one plus open x plus y close divided by x plus y

equation left hand side sum with, 3 , summands negative one plus one plus open x plus y close times d times open x plus y close divided by one plus open x plus y close equals right hand side two times d times x

equation left hand side d times open sum with, 3 , summands one plus x plus y close divided by one plus open x plus y close equals right hand side d times open y minus x close

natural log of absolute value of sum with, 3 , summands one plus x plus y equals open y minus x close plus natural log of cap c

equation left hand side sum with, 3 , summands one plus x plus y equals right hand side cap c times exp of y minus x

Hope this can be stretched to use in more complicated cases

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Valentin Fadeev

Facts about integrating factors

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Edited by Valentin Fadeev, Sunday, 18 Sep 2011, 23:26

Even if you are faced with a plain separable ODE, the process of separation of variables itself implies multiplying both parts by some factor. Thus the integrating factor seems to be one of the most devious tricks of solving equations.

There is a general path to establish its existence. It can be found in many textbooks. I am interested in some particular cases here which give beautiful solutions.

First, for a homogeneous equation it is possible to find a closed formula for the integrating factor.

It can be shown that for equation

cap m of x comma y times d times x plus cap n of x comma y times d times y equals zero ,

where M and N are homogeneous functions of their arguments integrating factor has the form:

mu equals one divided by x times cap m of x comma y plus y times cap n of x comma y

Apply this to equation:

open a squared times x minus y close times d times x plus open x plus y close times d times y equals zero

equation sequence mu equals one divided by x times open a squared times x minus y close plus y times open x plus y close equals one divided by a squared times x squared plus y squared

Multiplying both parts by this expression we obtain:

a squared times x minus y divided by a squared times x squared plus y squared times d times x plus x plus y divided by a squared times x squared plus y squared times d times y equals zero

Rearranging:

x times d times y minus y times d times x divided by a squared times x squared plus y squared plus a squared times x times d times x plus y times d times y divided by a squared times x squared plus y squared equals zero

d of x divided by y divided by a squared postfix plus left parenthesis x divided by y times right parenthesis squared plus one divided by two times d times open a squared times x squared plus y squared close divided by a squared times x squared plus y squared equals zero

And the result becomes obvious.

For the next example it is useful to note the fact that if mu is an integrating factor for equation cap m times d times x plus cap n times d times y equals zero giving solution in the form cap u of x comma y equals cap c , then equation left hand side mu sub one equals right hand side mu times phi of cap u where phi is any differentiable function shall also be an integrating factor. Indeed

equation sequence mu sub one times open cap m times d times x plus cap n times d times y close equals phi of cap u times mu times open cap m times d times x plus cap n times d times y close equals phi of cap u times d times cap u

giving the differential for the function normal cap phi of cap u equals integral phi of cap u d cap u

This leads to the following practical trick of finding the factor. All terms of the equations are split in two groups for each of which it is easy to find the integrating factor. Then each factor is written in the most general form involving an arbitrary function as described above. Finally we try to find such functions that make both factors identical.

Consider the following equation:

open x squared times y squared minus one close times d times y plus two times x times y cubed times d times x equals zero

Rearranging the terms:

open x squared times y squared times d times y plus two times x times y cubed times d times x close minus d times y equals zero

For the second term now the integrating factor is trivial, it is 1. Hence the most general form will look like mu sub one equals phi of y .

For the first part it is easy to see that the factor should be one divided by x squared times y cubed giving solution x squared times y equals cap c , hence equation left hand side mu sub two equals right hand side one divided by x squared times y cubed times psi of x squared times y .

To make the two identical we want mu sub two to be independent of x. Setting psi of t equals t gives equation left hand side mu sub two equals right hand side one divided by y squared .

Applying this one we get:

open x squared times d times y plus two times x times y times d times x close minus d times y divided by y squared equals zero

x squared times y plus one divided by y equals cap c

Both methods were discovered in the classic book "A course on differential equations" by V.V. Stepanov

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Valentin Fadeev

Generalized homogeneous equations

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Edited by Valentin Fadeev, Sunday, 16 Jan 2011, 23:20

One of the rarely used methods of solving ODEs applies to the so-called generalized homogeneous equations. The word "generalized" means that the terms are not homogeneous in the classic sense, if all variables are assigned the same dimension. But they may be made homogeneous in a wider sense by choosing the appropriate dimension for the dependent variable. Here is one example.

equation left hand side x cubed times d times y divided by d times x equals right hand side y times open x squared plus y close

If we assign dimension 1 to x and dx and dimension m to y and dy, then the left side has dimension 3+m-1=m+2 on the right side we have m+2 and 2m. To balance things let m+2=2m, hence m=2 and we get a "generalized homogeneous equation" of the 4th order. The trick is to let:

x equals e super t comma y equals u times e super m times t

which in this case gives:

equation left hand side d times x equals right hand side e super t times d times t comma equation left hand side d times y equals right hand side e super two times t times open u super prime plus two times u close

equation left hand side d times y divided by d times x equals right hand side e super t times open u super prime plus two times u close

Hence the equation becomes:

equation left hand side e super four times t times open u super prime plus two times u close equals right hand side u times e super four times t times open one plus u close

equation left hand side u super prime plus u equals right hand side u squared

u super prime divided by u squared plus one divided by u equals one

letting z=1/y

equation left hand side d times z divided by d times x equals right hand side z minus one

z equals one plus c times x

y equals x squared divided by one plus c times x

This method can of course, be applied to higher order equations

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