Author Topic: TUT0401 Solution  (Read 545 times)

EroSkulled

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TUT0401 Solution
« on: September 27, 2019, 02:00:00 PM »
Find general solution for $xy'=(1-y^2)^\frac{1}{2}$
$x\frac{dy}{dx}=\sqrt{1-y^2}$

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

Note this is a separable function, rearrange:
$\int{\frac{1}{\sqrt{1-y^2}}dy}=\int{\frac{1}{x}dx}$    where $x≠0, y≠±1$

Integrate both side:
$LHS: \int{\frac{1}{\sqrt{1-y^2}}dy}=\arcsin{y}$

$RHS: \int{\frac{1}{x}dx}=\ln{|x|}+C$

$arcsin(y)=\ln{|x|}+C$
∴General Solution is the following:

$y=\sin{(\ln{|x|}+C)}$     where $x≠0, y≠±1$