ydx = x + √ y^2 - x^2 dy
Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem.
ydx= x + ( y^2 - x^2) ^ 1/2 dy
The x don't have differential element, do you mean this? $y \, dx = x + (y^2 - x^2)^{1/2} \, dy$
Or this? $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$
ydx = [x + ( y^2 - x^2 )^ 1/2] dy
this one.
Solution $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$
$y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$
$vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$
$y^2\,dv = y(1 - v^2)^{1/2}\,dy$
$y\,dv = (1 - v^2)^{1/2}\,dy$
$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$
$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$
$\arcsin v = \ln y + c$
$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$
sa factoring po. bakit po naging.
[ vy + y (1- v^2 )^1/2 ] dy
(y2 - v2 y2)1/2 = [ y2 (1 - v2) ] 1/2 = y2(1/2) (1 - v2)1/2 = y (1 - v2)1/2
salamat po.
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Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem.
ydx= x + ( y^2 - x^2) ^ 1/2 dy
The x don't have differential element, do you mean this?
$y \, dx = x + (y^2 - x^2)^{1/2} \, dy$
Or this?
$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$
ydx = [x + ( y^2 - x^2 )^ 1/2] dy
this one.
Solution
$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$
dx = v dy + y dv
$y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$
$vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$
$y^2\,dv = y(1 - v^2)^{1/2}\,dy$
$y\,dv = (1 - v^2)^{1/2}\,dy$
$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$
$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$
$\arcsin v = \ln y + c$
$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$
sa factoring po. bakit po naging.
[ vy + y (1- v^2 )^1/2 ] dy
(y2 - v2 y2)1/2
= [ y2 (1 - v2) ] 1/2
= y2(1/2) (1 - v2)1/2
= y (1 - v2)1/2
salamat po.
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