Solving Differential Equation: Substitution Method

  • Thread starter -Vitaly-
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In summary, the conversation involves solving a differential equation using substitution and implicit solution. The solution involves using arctan and logarithmic functions.
  • #1
-Vitaly-
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Homework Statement


Solve: http://img384.imageshack.us/img384/60/clipboard01az7.jpg [Broken]

Homework Equations


DE stuff

The Attempt at a Solution


I started by doing substitution: x=U+3/2 and y=W-1/2

So, it gave me http://img522.imageshack.us/img522/3172/clipboard02ly8.jpg [Broken],[/URL] but http://img167.imageshack.us/img167/8297/clipboard03db9.jpg [Broken],[/URL] so http://img175.imageshack.us/img175/3223/clipboard04xt2.jpg [Broken].[/URL]

Therefore, http://img237.imageshack.us/img237/2264/clipboard05iu8.jpg [Broken]

And, http://img377.imageshack.us/img377/615/clipboard06ez2.jpg [Broken].[/URL] Substitution z=W/U. Giving, http://img139.imageshack.us/img139/5639/clipboard07hb7.jpg [Broken] (*).

And, http://img399.imageshack.us/img399/2373/clipboard08qj2.jpg [Broken]

Rearranging gives http://img114.imageshack.us/img114/1830/clipboard09tm2.jpg [Broken].[/URL] Substitute in (*)

http://img140.imageshack.us/img140/8541/clipboard10bz2.jpg [Broken] => http://img140.imageshack.us/img140/6706/clipboard11ao0.jpg [Broken] => http://img135.imageshack.us/img135/8163/clipboard12ee3.jpg [Broken]

Dividing 1/(1+z2) and -z/(1+z2) and integrating gives:

http://img521.imageshack.us/img521/3638/clipboard13nj5.jpg [Broken].[/URL] But now I'm stuck :( because I can't solve this for z and go back to y and x :( maybe I made an arithmetical mistake or maybe there is a better method?
Thank you
 
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  • #2
I don't know how correct the method you detailed is but a lot of times for a differential equation, an implicit solution is sufficient.
 
  • #3
Surely there is a way to solve it explicitely
 
  • #4
Why "surely". You don't need to solve
[tex]arctan z-\frac{1}{2}ln(z^2+ 1)= ln(U)+ C[/tex]

z= W/U so that is
[tex]arctan W/U-\frac{1}{2}ln(\frac{W^2}{U^2}+ 1)= ln(U)+ C[/tex]
and U= x- 3/2, W= y+ 1/2 so
[tex]arctan\frac{y+ 1/2}{x-3/2}- \frac{1}{2}ln(\frac{(y+1/2)^2}{(x-3/2)^2}= ln(x- 3/2)+ C[/tex]
That's a perfectly good solution. (Assuming your integration was correct.)
 

1. What is the substitution method for solving differential equations?

The substitution method involves replacing one or more variables in a differential equation with a new variable or expression. This is done in order to simplify the equation and make it easier to solve.

2. When is the substitution method used for solving differential equations?

The substitution method is typically used when the differential equation is not in a standard form and cannot be solved using other methods such as separation of variables or integrating factors. It is also useful when the equation contains complex or difficult-to-integrate terms.

3. How do you perform the substitution method for solving differential equations?

To perform the substitution method, you must first identify the variable to be substituted. Then, you must choose a new variable or expression to replace it with. Finally, you must substitute the new variable into the original equation and solve for it.

4. What are the advantages of using the substitution method for solving differential equations?

The substitution method can help simplify complex differential equations and make them easier to solve. It also allows for the use of different techniques and strategies to solve equations that may not be solvable using other methods.

5. Are there any limitations to using the substitution method for solving differential equations?

One limitation of the substitution method is that it may not always be possible to find a suitable substitution that will make the equation easier to solve. In addition, the process of substitution may introduce new variables or constants that make the solution more complex.

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