Integral Help: Solving dy/(4-y^0.5)

[kvanr]

Just wondering if you can help me out, I have the following integral that I go to in some differential equation I came up with to figure something out here:

Integrate: dy/(4-y^0.5)

The root y screws me up.

 

[Hurkyl]

The root y screws me up.

Then get rid of it. (With some sort of substitution)

 

[kvanr]

Then get rid of it. (With some sort of substitution)

Yah, how?
That's my question.

 

[Pyrrhus]

Try

$$z^{2} = y^{0.5}$$

$$dy = 4z^3 dz$$

 

[kvanr]

Try

$$z^{2} = y^{0.5}$$

$$dy = 4z^3 dz$$

Ah yes, interesting. So then I get it in the form

$$4z^3 dz / (4 - z^2)$$

I see you can simplify the $$4 - z^2 to:<br /> (2-z)(2+z)$$, but then I tried to do partial fractions, and that also went nowhere, as there is no existing way to do a partial fraction for that that I could find?

and again I am stumped.. sorry, I'm just not getting this.

 

[Hurkyl]

Cyclovenom: Why not just z = y^0.5? (or z^2 = y)

but then I tried to do partial fractions, and that also went nowhere, as there is no existing way to do a partial fraction for that that I could find?

This is a routine application of partial fractions. Maybe you forgot about a step? (the way I learned it... the first step)

 

[kvanr]

Cyclovenom: Why not just z = y^0.5? (or z^2 = y)


This is a routine application of partial fractions. Maybe you forgot about a step? (the way I learned it... the first step)

$$A/(2-x) + B/(2+x) = 4x^3/(4-x^2)$$
$$A(2+x) + B(2-x) = 4x^3$$
$$(2A+2B) + x(A-B) = 4x^3 + 0x + 0$$
2A+2B = 0, so A=-B
x(A-B)=0x, so A=B

Therefore, no solution exists by the method I know.

 

[Hurkyl]

$$A/(2-x) + B/(2+x) = 4x^3/(4-x^2)$$

There's something you're supposed to do before this...

(if you can't find it in your book, the answer is: you're supposed to divide first[/color])

 

[Pyrrhus]

Cyclovenom: Why not just z = y^0.5? (or z^2 = y)

Yes, it's easier, i guess i am gettin' rusty.

 

[kvanr]

There's something you're supposed to do before this...

(if you can't find it in your book, the answer is: you're supposed to divide first[/color])

I'm sorry I can't find it.

It's because I'm stupid.

edit: Long division.. hmm..

would it be the same as:
$$-4x + 16x / (4-x^2)$$

I checked and looks fine.. integrating now
so final answer of $$-2z^2 - 8ln(4-z^2)$$
Then just sub in root y for z^2