HOMOGENEOUS DE: $(x - y \ln y + y \ln x) dx + x(\ln y - \ln x) dy = 0$

Submitted by Sydney Sales on

1. (x - ylny + ylnx) dx + x(lny - lnx) dy= 0
2. (x csc y/x - y) dx + xdy=0
3. (x^2 + 2xy - 4y^2) dx - ( x^2 - 8xy - 4 y^2)=0
4. x^y ' = 4x^2 + 7xy + 2 y^2

Tags
Differential Equation, DE

Jhun Vert Fri, 07/08/2016 - 09:58

Solution to No. 1
$(x - y \ln y + y \ln x) \, dx + x(\ln y - \ln x) \, dy = 0$

$\big[ x - y(\ln y - \ln x) \big] \, dx + x(\ln y - \ln x) \, dy = 0$

$\left[ x - y\ln \left( \dfrac{y}{x} \right) \right] dx + x\ln \left( \dfrac{y}{x} \right) dy = 0$

Let
y = vx
dy = v dx + x dv

$\left[ x - vx\ln \left( \dfrac{vx}{x} \right) \right] dx + \left[ x\ln \left( \dfrac{vx}{x} \right) \right](v \, dx + x \, dv) = 0$

$(x - vx\ln v) \, dx + (x\ln v)(v \, dx + x \, dv) = 0$

$x \, dx - vx\ln v \, dx + vx\ln v \, dx + x^2 \ln v \, dv = 0$

$x \, dx + x^2 \ln v \, dv = 0$

$\dfrac{dx}{x} + \ln v \, dv = 0$

$\ln x + (v \ln v - v) = c$

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