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

Submitted by Sydney Sales on July 4, 2016 - 9:16am
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
- Add new comment
- 11832 reads
pasagot po...
pasagot po...
ano po ang solution....
ano po ang solution....
Solution to No. 1
Solution to No. 1
$(x - y \ln y + y \ln x) \, dx + x(\ln y - \ln x) \, dy = 0$
$\big[ x - y(\ln y - \ln x) \big] \, dx + x(\ln y - \ln x) \, dy = 0$
$\left[ x - y\ln \left( \dfrac{y}{x} \right) \right] dx + x\ln \left( \dfrac{y}{x} \right) dy = 0$
y = vx
dy = v dx + x dv
$\left[ x - vx\ln \left( \dfrac{vx}{x} \right) \right] dx + \left[ x\ln \left( \dfrac{vx}{x} \right) \right](v \, dx + x \, dv) = 0$
$(x - vx\ln v) \, dx + (x\ln v)(v \, dx + x \, dv) = 0$
$x \, dx - vx\ln v \, dx + vx\ln v \, dx + x^2 \ln v \, dv = 0$
$x \, dx + x^2 \ln v \, dv = 0$
$\dfrac{dx}{x} + \ln v \, dv = 0$
$\ln x + (v \ln v - v) = c$
$\ln x + \dfrac{y}{x} \ln \left( \dfrac{y}{x} \right) - \dfrac{y}{x} = c$
pano po ung problem 2,3,4 for
pano po ung problem 2,3,4 for homogeneous DE
how about the problem no.2,3
how about the problem no.2,3 and4
naka graduate kana sir?
naka graduate kana sir?
Add new comment