# Simple integral, that I don't understand

1. Jan 13, 2007

### snowJT

1. The problem statement, all variables and given/known data

I know how to integrate, but I don't understand the wording of the question and what it all means...

Question: How to get a particular solution from the general solution using a BOUNDARY CONIDTION.

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

subject to the condition that $$y = 2$$ when $$y = 3$$

2. The attempt at a solution

$$\frac{y^2}{2} = \frac{x^3}{3} + C$$

$$y = 2$$ and $$y = 3$$

replace into equation...

$$2 = 9+ C$$
$$C = -7$$

then I guess I replace it into the general solution to verify??

$$\frac{y^2}{2} = \frac{x^3}{3} + C$$

$$y^2 = \frac{2x^3}{3} - (2)7$$

$$y^2 = \frac{2(3)^3}{3} - (2)7$$

$$y^2 = \frac{2(3)^3}{3} - 14$$

$$y^2 = 18 - 14$$

$$y = \sqrt{4}$$

$$y = 2$$

I know how to integrate ect.. It's just I don't understand what the question wants? Maybe I solved it?

2. Jan 13, 2007

### Dick

The form with the C in it is the general form. You found the particular form when you determined C=-7. You are done.

3. Jan 13, 2007

### HallsofIvy

Staff Emeritus
Surely you mean "y= 2 when x= 3".

Again, the second equation is x= 3

Yes, that's exactly right.

This is correct.

Now I am confused. Why are you setting x equal to 3?

Yes, just like you were told, "when x= 3, y= 2"! Were you checking? The solution to the problem is $$y^2= \frac{2x^3}{3}- 14$$

Last edited: Jan 13, 2007
4. Jan 13, 2007

### Dick

I assumed y=3 was a typo. If the phrase "BOUNDARY CONIDTION" is the source of your confusion, then "x=3, y=2" IS a boundary condition. And you are done.

5. Jan 13, 2007

### snowJT

what else should I do with it?

ya sorry its suppose to read x = 3 then i copy pasted my mistake

6. Jan 13, 2007

### cristo

Staff Emeritus
you don't need to set x=3. This is a boundary condition, which you used to obtain the value of the constant of integration. You have done this, so have the solution $$y^2= \frac{2x^3}{3}- 14$$

7. Jan 13, 2007

### snowJT

you mean $$y^2= \frac{2x^3}{3}- 7$$ right? because it would be 14 if I made y = 2

8. Jan 13, 2007

### cristo

Staff Emeritus
No, i mean $$y^2= \frac{2x^3}{3}- 14$$(*). You had the equation $$\frac{y^2}{2} = \frac{x^3}{3} + C$$, which you solved for C to get C=-7. But now multiplying through by 2, we obtain (*)

9. Jan 13, 2007

### snowJT

oh... I missed that, I'm sorry

I don't think I'll ever bebale to catch you on anything, you're eye is too good

10. Jan 13, 2007

### Dick

Let's face it. His first post was correct except for typos in the posting. He just thought he might be supposed to do something else. Now he's confused. The phrase "Maybe I solved it?" should have simply been answered "yes".