Evaluating integral

  • Thread starter Saitama
  • Start date
  • #1
3,812
92

Homework Statement


If ##y(x-y)^2=x##, then find
$$\int \frac{dx}{x-3y}$$


Homework Equations





The Attempt at a Solution


I tried expanding the given equation. That gives me a cubic in y. I found no way to solve the cubic equation. I plugged the equation in Wolfram Alpha to see if it gives any nice solution but nope, no luck there too. Now I think that this is some sort of trick question but I don't really know where to begin.

Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
What about changing to the variable y/x=t? Just a guess...

ehild
 
  • #3
3,812
92
What about changing to the variable y/x=t? Just a guess...

ehild
I should have really attempted the problem properly before posting it here. I just woke up and gave it a try. I substituted x-y=t. The integral changes to dx/(3t-2x) and the equation changes to ##(x-t)t^2=x##. I differentiated the equation wrt x and after some simplification, I got
$$\frac{dx}{3t-2x}=\frac{dt}{t^2-1}$$

Now its a very easy integral to solve. Thanks for the effort ehild. :smile:
 
  • #4
1,796
53
I should have really attempted the problem properly before posting it here. I just woke up and gave it a try. I substituted x-y=t. The integral changes to dx/(3t-2x) and the equation changes to ##(x-t)t^2=x##. I differentiated the equation wrt x and after some simplification, I got
$$\frac{dx}{3t-2x}=\frac{dt}{t^2-1}$$

Now its a very easy integral to solve. Thanks for the effort ehild. :smile:
I think that's wrong. Sides, once you got t, then what? How about checking it with the integral:

[tex]\int_1^2 \frac{1}{x-3y}dx,\quad y(x-y)^2=x[/tex]

While we're at it, what is the geometric interpretation of that integral anyway?
 
  • #5
3,812
92
I think that's wrong. Sides, once you got t, then what?
I don't see what's wrong here. I differentiated both the sides of the equation wrt x. Did I do some mistake while simplifying the expression?
 
  • #6
3,812
92
Just realised that the final equation after differentiating I get is wrong. It is actually
$$\frac{dx}{3t-2x}=\frac{t \, dt}{t^2-1}$$
 
  • #7
1,796
53
Just realised that the final equation after differentiating I get is wrong. It is actually
$$\frac{dx}{3t-2x}=\frac{t \, dt}{t^2-1}$$
To check it, I solved the definite integral above and obtained:

[tex]\int_1^2 \frac{1}{x-3y} dx=\int_a^b \frac{t}{t^2-1}dt[/tex]

where a=1-lowroot and b=2-highroot with the roots being the real roots of the equations [itex]y(1-y)^2=1[/itex] and [itex]y(2-y)^2=2[/itex]

which agrees numerically with direct substitution of y(x) from the solution of [itex]y(x-y)^2=x[/itex].

Always a good idea to check them numerically if we're trying to intercept an earth-bound meteor.
 
  • #8
21
1
Given ::##y\cdot(x-y)^2 = x##


Now Diff. both side w. r . to ##x## , We Get


##\displaystyle y\cdot 2(x-y)\left\{1-\frac{dy}{dx}\right\}+(x-y)^2 \cdot \frac{dy}{dx} = 1##


##\displaystyle 2y\cdot(x-y)\left\{1-\frac{dy}{dx}\right\}+\frac{dy}{dx}\cdot (x-y)^2-(x-y)^2 = 1-(x-y)^2##


##\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(2xy-y^2-x^2+2xy-y^2\right) = 1-(x-y)^2##


##\displaystyle \left\{1-\frac{dy}{dx}\right\}\left(-x^2+4xy-3y^2\right) = 1-(x-y)^2##


##\displaystyle \left\{\frac{dy}{dx}-1 \right\}\left(x^2-4xy+3y^2\right) = 1 - (x-y)^2##


##\displaystyle \left\{\frac{dy}{dx}-1 \right\}.\left(x-y\right).\left(x-3y\right) = 1-(x-y)^2##


##\displaystyle \left(dy - dx\right).(x-y).(x-3y) = \left\{1-(x-y)^2\right\}dx##


##\displaystyle \frac{dx}{(x-3y)} = \frac{\left(dy - dx\right)\cdot(x-y)}{1-(x-y)^2}##


Now Integrate both side , We Get


##\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{\left(dx - dy\right)\cdot(x-y)}{(x-y)^2 -1}##


Let ##(x-y) = t\Leftrightarrow d(x-y) = dt\Leftrightarrow \left(dx - dy\right) = dt##


So ##\displaystyle \int\frac{dx}{(x-3y)} = \int \frac{t}{t^2-1}dt##


So again Let ##\displaystyle t^2 - 1 = u\Leftrightarrow tdt = \frac{1}{2}du##


So ##\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \int\frac{1}{u}du = \frac{1}{2}\cdot \ln \left|u\right|+C##


So ##\displaystyle \int\frac{dx}{(x-3y)} = \frac{1}{2}\cdot \ln \left|(x-y)^2 - 1\right|+C##
 

Related Threads on Evaluating integral

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
698
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
3
Views
724
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
9
Views
836
  • Last Post
Replies
1
Views
2K
Top