Another Seperation of Variables Problem

[Jason76]

Looking at the same one from the other thread:

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

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

$\displaystyle 2\int \dfrac{dy}{(1 - y^{2})^{1/2}} = \int x^{-1/2}dx$ Did the 2 on the far left side come from substitution?

$\displaystyle 2 \arcsin y + C_y = 2x^{1/2} + C_x$

$\displaystyle 2 \arcsin y + C_y - C_y - = 2x^{1/2} + C_x - C_y$

$\displaystyle 2 \arcsin y = 2x^{1/2} + C$

$\displaystyle \arcsin y = x^{1/2} + C$

$\displaystyle y = \sin(\sqrt{x} + C)$

 

[Deleted member 4993]

Looking at the same one from the other thread:

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

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

$\displaystyle 2\int \dfrac{dy}{(1 - y^{2})^{1/2}} = \int x^{-1/2}dx$ Did the 2 on the far left side come from substitution?

What substitution?

You had a "2" on the far-left-side from the first line down....

$\displaystyle 2 \arcsin y + C_y = 2x^{1/2} + C_x$

$\displaystyle 2 \arcsin y + C_y - C_y - = 2x^{1/2} + C_x - C_y$

$\displaystyle 2 \arcsin y = 2x^{1/2} + C$

$\displaystyle \arcsin y = x^{1/2} + C$

$\displaystyle y = \sin(\sqrt{x} + C)$

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[Jason76]

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So all constants of integration don't come from substitution.

How does this problem relate to the integral: $\displaystyle \int \arcsin x dx = x \arcsin x + (1 - x^{2})^{1/2} + C$ ? I can see somehow how it relates.

 

[Deleted member 4993]

So all constants of integration don't come from substitution. .... I am not sure what do you mean by that.

How does this problem relate to the integral: $\displaystyle \int \arcsin x dx = x \arcsin x + (1 - x^{2})^{1/2} + C$ ? I can see somehow how it relates.

I don't see any relation - but then relation is in the eye of the beholder.

$\displaystyle \int \arcsin x dx = x \arcsin x + (1 - x^{2})^{1/2} + C$

use integration by parts:

u = sin^-1(x)

dv = dx