Help solving the initial value problem

[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? I am 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.

 

[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?

 

[Saeed.z]

let

and

find the integrating factor

and then use

hope that help :)

 

[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

 

[Saeed.z]

you firstly have to integrate the right side

before you multiply it by

!

 

[middieman147]

Ok:

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

 

[Saeed.z]

when u divide the right side after integrating by

, u should also divide the constant too so it should be

 

[HallsofIvy]

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

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

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? I am 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.

 

[middieman147]

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

yes I do.


thanks.

 

[middieman147]

when u divide the right side after integrating by

, u should also divide the constant too so it should be

wouldn't that just be another constant?

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

 

[SammyS]

wouldn't that just be another constant?

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

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

 

[middieman147]

isnt that irrelevant since there is an IC? wouldn't 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?

 

[SammyS]

isnt that irrelevant since there is an IC? wouldn't 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?

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.