# Help solving the initial value problem

1. Mar 25, 2012

### middieman147

Given:
Solve the initial value problem 2(√x)y'+y+4(√x) ; y(1)=2

I am having trouble separating the x's and y's in order to integrate. I keep coming up with:

dy/dx +y/(2(√x))=2.....

What do I keep missing here? Im pretty sure you leave the y(1)=2 alone until you are finished with the integration, in which case you plug in x=1 and y=2 to solve for the constant. Is this the correct thought process for this problem?

Thanks.

2. Mar 25, 2012

### middieman147

Or can I substitute in the y(1) for the 2 on the right side of the equation and make all the other x's equal to 1?

3. Mar 25, 2012

### Saeed.z

let

and

find the integrating factor

and then use

hope that help :)

4. Mar 25, 2012

### middieman147

Ok so now I have:

y'-y/(2√(x))=2

p(x)=-1/(2√x) q(x)=2

u(x)=e∫(-1/(2√x))=e-√x

y=e√x(2e-√x)+C

y=2+C

this is the solution for all values of x with C=0, correct? Or am I forgetting a step?

Thanks

5. Mar 25, 2012

### Saeed.z

you firstly have to integrate the right side before you multiply it by !

6. Mar 25, 2012

### middieman147

Ok:

y=-4sqrt(x)-4+C with C=10 is the answer given the IC y(1)=2?

7. Mar 25, 2012

### Saeed.z

when u divide the right side after integrating by , u should also divide the constant too so it should be

8. Mar 25, 2012

### HallsofIvy

Staff Emeritus
Do you mean $2\sqrt{x}y'+ y= 4\sqrt{x}$?

9. Mar 25, 2012

### middieman147

yes I do.

thanks.

10. Mar 25, 2012

### middieman147

wouldn't that just be another constant?

so instead of writing Ce^sqrt(x), you could just consider it a new constant

11. Mar 25, 2012

### SammyS

Staff Emeritus
C is a constant, but $Ce^\sqrt{x}$ is definitely not a constant.

12. Mar 25, 2012

### middieman147

isnt that irrelevant since there is an IC? wouldnt the x in that expression =1 and then just equal C multiplied by another constant?

edit: Are you saying i should divide 10 by e^1 to find the constant?

13. Mar 26, 2012

### SammyS

Staff Emeritus
No, that's not irrelevant.

In using the Initial Condition to evaluate the constant, C , you do set x=1, but in the overall solution there is a term $Ce^\sqrt{x}$, unless the Initial Condition gives C=0, which is not the case here.