Differential EQNS: $y \, dx = \left[ x + (y^2 - x^2)^{1/2} \right] dy$ Submitted by Sydney Sales on Thu, 07/14/2016 - 10:53 ydx = x + √ y^2 - x^2 dy Tags Differential Equation, DE Log in to post comments Please use proper groupings Jhun Vert Thu, 07/14/2016 - 15:49 Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem. Log in to post comments ydx= x + ( y^2 - x^2) ^ 1/2 Sydney Sales Thu, 07/14/2016 - 21:41 ydx= x + ( y^2 - x^2) ^ 1/2 dy Log in to post comments The x don't have differential Jhun Vert Fri, 07/15/2016 - 05:26 The x don't have differential element, do you mean this? $y \, dx = x + (y^2 - x^2)^{1/2} \, dy$ Or this? $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$ Log in to post comments ydx = [x + ( y^2 - x^2 )^ 1/2 Sydney Sales Fri, 07/15/2016 - 07:30 ydx = [x + ( y^2 - x^2 )^ 1/2] dy this one. Log in to post comments Solution Jhun Vert Fri, 07/15/2016 - 12:46 Solution $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$ Let x = vy dx = v dy + y dv $y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$ $vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$ $y^2\,dv = y(1 - v^2)^{1/2}\,dy$ $y\,dv = (1 - v^2)^{1/2}\,dy$ $\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$ $\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$ $\arcsin v = \ln y + c$ $\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$ Log in to post comments sa factoring po. bakit po Sydney Sales Fri, 07/15/2016 - 19:08 sa factoring po. bakit po naging. [ vy + y (1- v^2 )^1/2 ] dy Log in to post comments (y2 - v2y2)1/2 Jhun Vert Fri, 07/15/2016 - 23:51 In reply to sa factoring po. bakit po by Sydney Sales(y2 - v2 y2)1/2 = [ y2 (1 - v2) ] 1/2 = y2(1/2) (1 - v2)1/2 = y (1 - v2)1/2 Log in to post comments salamat po. Sydney Sales Sat, 07/16/2016 - 14:13 salamat po. Log in to post comments
Please use proper groupings Jhun Vert Thu, 07/14/2016 - 15:49 Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem. Log in to post comments
ydx= x + ( y^2 - x^2) ^ 1/2 Sydney Sales Thu, 07/14/2016 - 21:41 ydx= x + ( y^2 - x^2) ^ 1/2 dy Log in to post comments
The x don't have differential Jhun Vert Fri, 07/15/2016 - 05:26 The x don't have differential element, do you mean this? $y \, dx = x + (y^2 - x^2)^{1/2} \, dy$ Or this? $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$ Log in to post comments
ydx = [x + ( y^2 - x^2 )^ 1/2 Sydney Sales Fri, 07/15/2016 - 07:30 ydx = [x + ( y^2 - x^2 )^ 1/2] dy this one. Log in to post comments
Solution Jhun Vert Fri, 07/15/2016 - 12:46 Solution $y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$ Let x = vy dx = v dy + y dv $y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$ $vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$ $y^2\,dv = y(1 - v^2)^{1/2}\,dy$ $y\,dv = (1 - v^2)^{1/2}\,dy$ $\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$ $\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$ $\arcsin v = \ln y + c$ $\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$ Log in to post comments
sa factoring po. bakit po Sydney Sales Fri, 07/15/2016 - 19:08 sa factoring po. bakit po naging. [ vy + y (1- v^2 )^1/2 ] dy Log in to post comments
(y2 - v2y2)1/2 Jhun Vert Fri, 07/15/2016 - 23:51 In reply to sa factoring po. bakit po by Sydney Sales(y2 - v2 y2)1/2 = [ y2 (1 - v2) ] 1/2 = y2(1/2) (1 - v2)1/2 = y (1 - v2)1/2 Log in to post comments
Please use proper groupings
Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem.
ydx= x + ( y^2 - x^2) ^ 1/2
ydx= x + ( y^2 - x^2) ^ 1/2 dy
The x don't have differential
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
ydx = [x + ( y^2 - x^2 )^ 1/2] dy
this one.
Solution
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
sa factoring po. bakit po naging.
[ vy + y (1- v^2 )^1/2 ] dy
(y2 - v2y2)1/2
In reply to sa factoring po. bakit po by Sydney Sales
(y2 - v2 y2)1/2
= [ y2 (1 - v2) ] 1/2
= y2(1/2) (1 - v2)1/2
= y (1 - v2)1/2
salamat po.
salamat po.