Differential equation: 4sqrt(xy)dy/dx=1, y(1)=1

[Lo.Lee.Ta.]

The differential equation I'm working on is:

4(√(xy))dy/dx=1, y(1)=1

(√(xy)dy/dy)² = (1/4)²

((xy)dy/dx)*dx = (1/8)*dx

(xy)dy = (1/8)dx

(y)dy = (1/(8x))dx

...So I think this is right so far.

Now I'm going to take the integral of both sides.

∫(y)dy = ∫(1/(8x))dx <------∫(1/x)(1/8)dx

1/2y² = ln|x|*(1/8) + C

√[y²] = √[2(ln|x|*(1/8) + C)]

y= √(1/4*ln|x| + C)

...Substitution in the y(1)=1

1 = √(1/4*ln|1| + C)

4 = 0 + C

C = 4

y(x)= √(1/4*ln|x| + 4)

This is counted as the wrong answer, but I don't know what I'm doing wrong here... #=_=
Thank you so much!

 

[Dick]

The differential equation I'm working on is:

4(√(xy))dy/dx=1, y(1)=1

(√(xy)dy/dy)² = (1/4)²

((xy)dy/dx)*dx = (1/8)*dx

(xy)dy = (1/8)dx

(y)dy = (1/(8x))dx

...So I think this is right so far.

Now I'm going to take the integral of both sides.

∫(y)dy = ∫(1/(8x))dx <------∫(1/x)(1/8)dx

1/2y² = ln|x|*(1/8) + C

√[y²] = √[2(ln|x|*(1/8) + C)]

y= √(1/4*ln|x| + C)

...Substitution in the y(1)=1

1 = √(1/4*ln|1| + C)

4 = 0 + C

C = 4

y(x)= √(1/4*ln|x| + 4)

This is counted as the wrong answer, but I don't know what I'm doing wrong here... #=_=
Thank you so much!

Your doing a lot of stuff wrong. Don't square it. Just separate the variables.

 

[Lo.Lee.Ta.]

Would you please tell me where exactly I squared wrongly?

Do you mean in the very beginning- my 2nd step?

"(√(xy)dy/dy)² = (1/4)²"

If this is where you mean, isn't it necessary to square first in order to get rid of the square root and then isolate the x from the y?

 

[SteamKing]

dy and dx are not immune to being squared.

 

[Lo.Lee.Ta.]

Ha! Oh! ...I actually wasn't sure if they'd be squared or not! But I guess they ARE squared! Thanks! ;)

 

[Lo.Lee.Ta.]

So my 2nd step really should be: 4√(y)dy = 1/√(x)dx?


4*√(x)*√(y)dy/dx = 1

4√(y) dy = 1/√(x)dx

 

[Lo.Lee.Ta.]

ugh, I don't know.

I'm still not getting the right answer... =_=

If this is correct: ∫4√(y)dy = ∫1/(√(x))dx, it seems it should be worked out like this:

(2/3)(4y^3/2) = 2x^1/2 + C

= (8/3)y^3/2 = 2√(x) + C

= y^3/2 = 3/8(2√(x) + C)

= y= √((3/8(2√(x) + C))³)Solving for C:

(1)^(3/2) = (2sqrt(1) + C)*3/8

1 = .75 + 3/8C

.25 = 3/8C

C= 2/3Seems that f(x)= sqrt[(3/8(sqrt(x)) + 2/3)^3]...

But this is incorrect...

 

[haruspex]

As Dick said, separate the variables: get all the y terms on one side and all the x on the other. This includes dy and dx. Then you can just integrate.

 

[ehild]

ugh, I don't know.

I'm still not getting the right answer... =_=

If this is correct: ∫4√(y)dy = ∫1/(√(x))dx, it seems it should be worked out like this:

(2/3)(4y^3/2) = 2x^1/2 + C

= (8/3)y^3/2 = 2√(x) + C

= y^3/2 = 3/8(2√(x) + C)[STRIKE]
= y= √((3/8(2√(x) + C))³)[/STRIKE]Solving for C:
.

You made an error again. If y^3/2=A then y=A^2/3

And it is easiest to get C from the equation in bold.

ehild

 

[Lo.Lee.Ta.]

Thanks very much, ehild! :)
You showed me where exactly I went wrong!

y^3/2 = 3/8(2√(x) + C)

y= (3/8(2√(1) + C))^2/3

1 = 3/8(2 + C)

C= 2/3

f(x) = (3/8(2√(x) + 2/3))^2/3

So this is counted as the right answer! Thanks! :)