# Help in checking the solution of this separable equation

1. Jul 8, 2015

### enggM

1. The problem statement, all variables and given/known data
It is just an evaluation problem which looks like this dx/dy = x^2 y^2 / 1+x

2. Relevant equations
dx/dy = x^2 y^2 / 1 + x

3. 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.

2. Jul 8, 2015

### HallsofIvy

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. Jul 9, 2015

### enggM

oh, so when i integrate [1+x / x] dx would it look like ∫1/x + ∫x dx? or something else?

4. Jul 9, 2015

### LCKurtz

1 + x/x = 1+1 =2.

5. Jul 9, 2015

### Staff: Mentor

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).

6. Jul 10, 2015

### enggM

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. Jul 10, 2015

### Staff: Mentor

Yes, you can split $\int \frac{1 + x}{x} \ dx$ into $\int \frac {dx} x + \int 1 \ dx$.

8. Jul 11, 2015

### enggM

so the answer y^3 = 3x + 3 ln x + C should be correct? ok i get it now thanks for the time.

9. Jul 12, 2015

### Staff: Mentor

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. Jul 19, 2015

### enggM

@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. Jul 19, 2015

### Ray Vickson

From
$$y^2 dy = \left( 1 + \frac{x}{x^2} \right) dx$$
you will get
$$\frac{1}{3} y^3 = x + \ln (x) + C$$
From
$$y^2 dy = \frac{1 + x}{x^2} dx$$
you will get
$$\frac{1}{3} y^3 = -\frac{1}{x} + \ln (x) + C$$

Which do you mean? Why are you still refusing to use parentheses? Do you not see their importance?

12. Jul 19, 2015

### Staff: Mentor

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. Jul 28, 2015

### enggM

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. Jul 28, 2015

### SammyS

Staff Emeritus
What is $\displaystyle \int\frac{1}{x^2}dx\ ?$

15. Jul 29, 2015

### Mentallic

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. Aug 5, 2015

### enggM

@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. Aug 5, 2015

### LCKurtz

Apparently you haven't been paying attention to anything people in this thread have told you about the importance of using parentheses.