solve for homogeneous

ydx = [ x + (y^2 - x^2)^1/2] dy

pa help naman poh pls

$y \, dx = [ \, x + (y^2 - x^2)^{1/2} \, ] \, dy$

Let $x = vy$

$dx = v \, dy + y \, dv$

$y(v \, dy + y \, dv) = [ \, vy + (y^2 - v^2 y^2)^{1/2} \, ] \, dy$

$vy \, dy + y^2 \, dv = vy \, dy + [ \, y^2(1 - v^2) \, ]^{1/2} \, dy$

$y^2 \, dv = y(1 - v^2)^{1/2} \, dy$

$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{y \, dy}{y^2}$

$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$

$\arcsin v = \ln y + c$

$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$

More information about text formats

Follow @iMATHalino

MATHalino

$y \, dx = [ \, x + (y^2 - x^2)^{1/2} \, ] \, dy$

Let

$x = vy$

$dx = v \, dy + y \, dv$

$y(v \, dy + y \, dv) = [ \, vy + (y^2 - v^2 y^2)^{1/2} \, ] \, dy$

$vy \, dy + y^2 \, dv = vy \, dy + [ \, y^2(1 - v^2) \, ]^{1/2} \, dy$

$y^2 \, dv = y(1 - v^2)^{1/2} \, dy$

$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{y \, dy}{y^2}$

$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$

$\arcsin v = \ln y + c$

$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$

## Add new comment