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

• kvanr
In summary, the integral cannot be solved by the method that the person is using because they forgot to do a step.
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.

The root y screws me up.
Then get rid of it. (With some sort of substitution)

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

Yah, how?
That's my question.

Try

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

$$dy = 4z^3 dz$$

Cyclovenom said:
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 $$4 - z^2 to: (2-z)(2+z)$$, 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.

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

kvanr said:
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)

Hurkyl said:
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.

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

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

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

Hurkyl said:
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)

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

Last edited:

## 1. What is the purpose of integral help for solving dy/(4-y^0.5)?

The purpose of integral help is to find the antiderivative or integral of a given function, in this case dy/(4-y^0.5). This helps in solving various mathematical problems and understanding the behavior of a function.

## 2. How do I solve dy/(4-y^0.5) using integration?

To solve this integral, you can use the substitution method by letting u = 4-y^0.5. Then, dy = -0.5y^-0.5 du. Substituting these values into the integral, we get ∫ -0.5y^-0.5 du/u. This can be simplified to -∫du, which gives us the solution of -u + C. Replacing u with 4-y^0.5, the final solution is -4 + y^0.5 + C.

## 3. What is the importance of understanding integrals for solving dy/(4-y^0.5)?

Understanding integrals is crucial for solving various mathematical problems, as it allows us to find the area under a curve, calculate volumes, and determine the net change of a function. In this case, understanding integrals helps us solve for the antiderivative of dy/(4-y^0.5) and find its solution.

## 4. Can I use any other method besides substitution to solve dy/(4-y^0.5)?

Yes, there are other methods that can be used to solve this integral, such as integration by parts or partial fractions. However, the substitution method is the most straightforward and efficient method for this particular integral.

## 5. How can I check if my solution to dy/(4-y^0.5) using integration is correct?

You can check your solution by differentiating it and seeing if it gives you the original function, dy/(4-y^0.5). If it does, then your solution is correct. You can also use online tools or software to verify your answer.

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