Simple integral, that I don't understand

[snowJT]

Homework Statement

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?

 

[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.

 

[HallsofIvy]

Homework Statement

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

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

2. The attempt at a solution

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

y = 2 and y = 3

Again, the second equation is x= 3

replace into equation...

2 = 9+ C
C = -7

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

Yes, that's exactly right.

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

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

This is correct.

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

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

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

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?

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

 

[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.

 

[snowJT]

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

what else should I do with it?

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

 

[cristo]

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

 

[snowJT]

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

 

[cristo]

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 (*)

 

[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

 

[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".