Find general solution of 2sqrt(x)(dy/dx) = sqrt(1 - y^2)

Hello everybody I'm stuck on this problem and need some help with:

. . .2sqrt(x) (dy/dx) = sqrt(1 - y^2)

I was able to get this far:

. . .2sqrt(x)/dx = sqrt(1 - y^2)/dy

Now we have to integrate both sides. This is where I'm stuck. Thanks!!!!

Re: Find general solution

Hello, mcwang719!

You're doing something weird!

\(\displaystyle \sqrt{x}\left(\frac{dy}{dx}\right) \:= \:\sqrt{1\,-\,y^2}\)

I was able to get this far: \(\displaystyle \L\, \frac{2\sqrt{x}}{dx} \:= \:\frac{\sqrt{1\,-\,y^2}}{dy}\;\;\) ??

Click to expand...

How do you expect to integrate with the differentials in the denominator ?

Arrange it like this: \(\displaystyle \L\,\frac{2\,dy}{\sqrt{1\,-\,y^2}} \:=\:\frac{dx}{\sqrt{x}}\)

Then we have: \(\displaystyle \L\,2\int\frac{dy}{\sqrt{1\,-\,y^2}} \:=\:\int x^{-\frac{1}{2}}\,dx\)

Can you finish it now?

Thanks a lot Soroban! Yes I can finish it now. Yeah I was looking at it thinking how the heck am I suppose to integrate this!

Have you never taken Calculus? \(\displaystyle \int\frac{dy}{\sqrt{1- y^2}}\) can be found in any Calculus text's "table of integrals" or done "by hand" by the substitution \(\displaystyle y= sin(\theta)\). Of course, \(\displaystyle \int x^{1/2} dx\) can be done by \(\displaystyle \int x^n dx= \frac{1}{n+1}x^{n+1}+ C\).

As for "How do they derive to the answer y= c^-x^2" they don't. That is obviously NOT the correct answer. I suspect you have misread it.

Separate the variables (y to the left and x to the right) and then integrate.


\(\displaystyle 2\sqrt{x}\dfrac{dy}{dx} = \sqrt{(1 - y^2)}\)

Rewritten using 1/2 power.

\(\displaystyle 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}\)

\(\displaystyle (dx) 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}(dx)\)

\(\displaystyle 2(x)^{1/2} dy= (1 - y^{2})^{1/2}dx\)

\(\displaystyle (\dfrac{1}{x^{1/2}})2(x)^{1/2}dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle 2 dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle (\dfrac{1}{2})2dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})(\dfrac{1}{ (1^{1/2} - y)})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy =(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

What now?

Jason76 said:

Separate the variables (y to the left and x to the right) and then integrate.


\(\displaystyle 2\sqrt{x}\dfrac{dy}{dx} = \sqrt{(1 - y^2)}\)

Rewritten using 1/2 power.

\(\displaystyle 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}\)

\(\displaystyle (dx) 2(x)^{1/2}\dfrac{dy}{dx} = (1 - y^{2})^{1/2}(dx)\)

\(\displaystyle 2(x)^{1/2} dy= (1 - y^{2})^{1/2}dx\)

\(\displaystyle (\dfrac{1}{x^{1/2}})2(x)^{1/2}dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle 2 dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})dx\)

\(\displaystyle (\dfrac{1}{2})2dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1 - y^{2})^{1/2}(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy= (1^{1/2} - y)(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})(\dfrac{1}{ (1^{1/2} - y)})dx\)

\(\displaystyle (\dfrac{1}{ (1^{1/2} - y)})dy =(\dfrac{1}{x^{1/2}})(\dfrac{1}{2})dx\) ................. Incorrect

What now?

Click to expand...

It should be

\(\displaystyle \dfrac{dy}{\sqrt{1-y^2}} \ = \ \dfrac{dx}{2\sqrt{x}}\)

Final solution:

y = sin(√x + c)
.

Subhotosh Khan said:

It should be

\(\displaystyle \dfrac{dy}{\sqrt{1-y^2}} \ = \ \dfrac{dx}{2\sqrt{x}}\)

Final solution:

y = sin(√x + c)
.

Click to expand...

So was my method "a more difficult way" or a wrong way?

Jason76 said:

So was my method "a more difficult way" or a wrong way?

Click to expand...

I cannot tell whether your method was wrong or not - because you had not finished it.

If you had continued on, without making further mistakes, you would have gotten an incorrect answer - because your intermediate "algebra" was incorrect.

Subhotosh Khan said:

I cannot tell whether your method was wrong or not - because you had not finished it.

If you had continued on, without making further mistakes, you would have gotten an incorrect answer - because your intermediate "algebra" was incorrect.

Click to expand...

Posted a thread in "intermediate Algebra" dealing with this problem.