Determine which of the following differential equations are homogeneous, then

Cheexorgeny

Cheexorgeny

Answered question

2021-12-31

Determine which of the following differential equations are homogeneous, then solve the general solution. If initial condition is given, find the particular solution:
xydx(x+2y)2dy=0

Answer & Explanation

Jim Hunt

Jim Hunt

Beginner2022-01-01Added 45 answers

xydx(x+2y)2dy=0
xydx=(x+2y)2dy
dxdy=x2+4y2+4xyxy
dxdy=(xy)+4(yx)+4
It is a Homogeneous differential equation.
Let xy=vx=vy
So, dxdy=v+ydvdy
v+ydvdy=v+4v+4
ydvdy=4(v+1v)
vdvv+1=4dyy
It becomes variable separable equation. So, now integrate both sides.
vdvv+1=4dyy
(v+11v+1)dv=4dyy
(11v+1)dv=4ln|y|+lnc
vln|v+1|=ln(y4c)
xy+ln|xy+1|=ln|y4c|
xy=ln|y4c|ln|x+yy|
xy=ln|(y4c)(x+

Serita Dewitt

Serita Dewitt

Beginner2022-01-02Added 41 answers

xydx(x+2y)2dy=0
xydx=(x+2y)2dy
dxdy=(x+2y)2xy
dxdy=x2+4y2+4xyxy
dxdy=xy+4yx+4.
Substituting; x=yv
Diff. w.r.t. y
dxdy=v+ydvdy
v+ydvdy=v+41v+4
¬{v}+ydvdy=¬{v}+4(1v+1)
y4dvdy=1+vv
Integrating vv+1dv=41ydy
v+11v+1dv=41ydy
vln|v+1|=4logy+c
applying v=xy
xyln|xy+1|=4logy+c
karton

karton

Expert2022-01-10Added 613 answers

xydx(x+2y)2dy=0xydx=(x+2y)2dydxdy=x2+4y2+4xyxydxdy=(xy)+4(yx)+4xy=vx=vydxdy=v+ydvdyv+ydvdy=v+4v+4ydvdy=4(v+1v)vdvv+1=4dyyvdvv+1=4dyy(v+11v+1)dv=4dyy(11v+1)dv=4ln|y|+lncvln|v+1|=ln(y4c)xy+ln|xy+1|=ln|y4c|xy=ln|y4c|ln|x+yy|xy=ln|(y4c)(x+yy)|x=yln|cy5x+y|

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