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Help in checking the solution of this separable equation

  • Thread starter enggM
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  • #1
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Homework Statement


It is just an evaluation problem which looks like this dx/dy = x^2 y^2 / 1+x

Homework Equations


dx/dy = x^2 y^2 / 1 + x

The Attempt at a Solution


What i did is cross multiply to get this equation y^2 dy = x^2 / 1+x dx then next line ∫y^2 dy = ∫x^2/1+x dx

y^3/3 = ∫dx + ∫1/x dx after simplifying i get y^3=3x + 3 ln x + C but im not sure if this is the right answer.
 

Answers and Replies

  • #2
HallsofIvy
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First, I presume that you really mean dx/dy= x^2y^2/(1+ x). What you wrote, dx/dy= x^2y^2/1+ x, is the same as dx/dy = x^2y^2+ x.

More importantly you have separated the x and y incorrectly. The left side is y^2 dy but the right sides should be [(1+ x)/x]dx. You have the fraction inverted.
 
  • #3
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oh, so when i integrate [1+x / x] dx would it look like ∫1/x + ∫x dx? or something else?
 
  • #4
LCKurtz
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oh, so when i integrate [1+x / x] dx would it look like ∫1/x + ∫x dx? or something else?
1 + x/x = 1+1 =2.
 
  • #5
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oh, so when i integrate [1+x / x] dx would it look like ∫1/x + ∫x dx? or something else?
Or did you intend the above to mean (1 + x)/x? If that's what you intended, it does absolutely no good to put brackets around the entire expression.

In post #1 you wrote what I've quoted below. For that, you need parentheses around the denominator, as x2y2/(1 + x).

It is just an evaluation problem which looks like this dx/dy = x^2 y^2 / 1+x
.
 
  • #6
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Oh sorry about that, what i intend to do is to integrate the entire expression. As it is the right hand side of the equation, no problem with y^2 dy but the right side looks a bit confusing. In the expression (1 + x) / x dx as was suggested is what i intend to integrate, so is this the right integration expression, ∫1 / x dx + ∫ dx? if so then ln x + x + C should be sufficient for the RHS, is it not?
 
  • #7
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Oh sorry about that, what i intend to do is to integrate the entire expression. As it is the right hand side of the equation, no problem with y^2 dy but the right side looks a bit confusing. In the expression (1 + x) / x dx as was suggested is what i intend to integrate, so is this the right integration expression, ∫1 / x dx + ∫ dx? if so then ln x + x + C should be sufficient for the RHS, is it not?
Yes, you can split ##\int \frac{1 + x}{x} \ dx## into ##\int \frac {dx} x + \int 1 \ dx##.
 
  • #8
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so the answer y^3 = 3x + 3 ln x + C should be correct? ok i get it now thanks for the time.
 
  • #9
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so the answer y^3 = 3x + 3 ln x + C should be correct? ok i get it now thanks for the time.
No @enggM, this is not correct. The work you did before was incorrect, and my earlier response was based on that work. I think you need to start from the beginning.

$$\frac{dx}{dy} = \frac{x^2y^2}{1 + x}$$

If you multiply both sides by 1 + x, then divide both sides by ##x^2##, and finally, multiply both sides by dy, the equation will be separated. What do you get when you do this?
 
  • #10
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@Mark44 i would get (1+x / x^2 ) dx = y^2 dy so integrating the both sides y^3 / 3 = 1 / x + ln x so the final form would then be y^3 = 3/x + 3 ln x + C? no?
 
  • #11
Ray Vickson
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@Mark44 i would get (1+x / x^2 ) dx = y^2 dy so integrating the both sides y^3 / 3 = 1 / x + ln x so the final form would then be y^3 = 3/x + 3 ln x + C? no?
From
[tex] y^2 dy = \left( 1 + \frac{x}{x^2} \right) dx [/tex]
you will get
[tex] \frac{1}{3} y^3 = x + \ln (x) + C[/tex]
From
[tex] y^2 dy = \frac{1 + x}{x^2} dx [/tex]
you will get
[tex] \frac{1}{3} y^3 = -\frac{1}{x} + \ln (x) + C[/tex]

Which do you mean? Why are you still refusing to use parentheses? Do you not see their importance?
 
  • #12
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@Mark44 i would get (1+x / x^2 ) dx = y^2 dy so integrating the both sides y^3 / 3 = 1 / x + ln x so the final form would then be y^3 = 3/x + 3 ln x + C? no?
In writing (1 + x/x2) above, you are using parentheses, but aren't using them correctly (as Ray also points out). If you have a fraction where either the numerator or denominator (or both) has multiple terms, you need parentheses around the entire numerator or denominator, not around the overall fraction.

If you mean ##\frac{1 + x}{x^2}##, write it in text as (1 + x)/x^2, NOT as (1+x / x^2 ). What you wrote means ##1 + \frac x {x^2} = 1 + \frac 1 x##.
 
  • #13
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sorry about the confusion because sometimes when i solve something like this in paper i sometimes leave out the parenthesis. so the final form would be y^3 = -3/x + 3 ln x + C? by the way how did it become a negative? just an additional question.
 
  • #14
SammyS
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sorry about the confusion because sometimes when i solve something like this in paper i sometimes leave out the parenthesis. so the final form would be y^3 = -3/x + 3 ln x + C? by the way how did it become a negative? just an additional question.
What is ##\displaystyle \int\frac{1}{x^2}dx\ ?##
 
  • #15
Mentallic
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sorry about the confusion because sometimes when i solve something like this in paper i sometimes leave out the parenthesis
On paper you can freely write the fraction as it's normally portrayed, however, if you mean that you do sometimes write everything on a single line and still don't use parentheses, then you need to break out of that habit. Parentheses are crucial.
 
  • #16
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@SammyS oh i see i remember now it should be u^-2+1 / -2 +1.
@Mentallic ok thanks...
thanks for all of your help in checking for the solution and answer to this problem helped me understand it much better.
 
  • #17
LCKurtz
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@SammyS oh i see i remember now it should be u^-2+1 / -2 +1.
.
Apparently you haven't been paying attention to anything people in this thread have told you about the importance of using parentheses.
 

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