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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Solve the Differential Equation $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{ \frac{1 - y^2}{1 - x^2} } \end{align*}$
$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \sqrt{ \frac{1 - y^2}{1 - x^2} } \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\sqrt{ 1 - y^2 }}{\sqrt{1 - x^2} } \\ \frac{1}{\sqrt{1 - y^2}} \, \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\sqrt{1 - x^2} } \\ \int{ \frac{1}{\sqrt{1 - y^2}}\, \frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x} &= \int{\frac{1}{\sqrt{1 - x^2}} \, \mathrm{d}x} \\ \int{ \frac{1}{\sqrt{1 - y^2}} \, \mathrm{d}y} &= \arcsin{(x)} + C_1 \\ \arcsin{(y)} + C_2 &= \arcsin{(x)} + C_1 \\ \arcsin{(y)} &= \arcsin{(x)} + C \textrm{ where } C = C_1 - C_2 \\ y &= \sin{ \left[ \arcsin{(x)} + C \right] } \\ y &= \sin{ \left[ \arcsin{(x)} \right] } \cos{(C)} + \cos{ \left[ \arcsin{(x)} \right] } \sin{(C)} \\ y &= A\sin{ \left[ \arcsin{(x)} \right] } + B\,\cos{ \left[ \arcsin{(x)} \right] } \textrm{ where } A = \cos{(C)} \textrm{ and } B = \sin{(C)} \\ y &= A\,x + B\,\sqrt{ 1 - \left\{ \sin{ \left[ \arcsin{(x)} \right] } \right\} ^2 } \\ y &= A\,x + B\,\sqrt{1 - x^2 } \end{align*}$

Make note of my use of the Compound Angle Formula for sine, and the Pythagorean Identity.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
just south of canada
I think you should note that $A$ and $B$ are not independent constants.... it might be better to write:

$y = x\sqrt{1 - B^2} + B\sqrt{1 - x^2}$