How to solve homogenous differential equation \(\displaystyle{\left({x}+{y}\right)}^{{2}}{s}{p}{a}{c}{e}{\left.{d}{x}\right.}={x}{y}{s}{p}{a}{c}{e}{\left.{d}{y}\right.}\)? First

ICESSRAIPCELOtblt

ICESSRAIPCELOtblt

Answered question

2022-03-31

How to solve homogenous differential equation (x+y)2dx=xy dy?
First i find value of dydx=(x+y)2xyt     (i)
let y=vx
differentiating both sides: dydx=v+xdvdx     (ii)
I tried to solve using equation (i) and (ii) but I am stuck.
(x+y)2x=xydy

Answer & Explanation

Pubephenedsjq

Pubephenedsjq

Beginner2022-04-01Added 11 answers

EXPLANATION:
I'm skipping some initial steps as you've figured that out. As y=vxyx=v.
The original equation finally reduces to 
v+xdvdx=1v+v+2

xdvdx=1+2vv

vdv1+2v=dxx
Can you take it up from here?

memantangti17

memantangti17

Beginner2022-04-02Added 13 answers

Step 1
We are given that dydx=(x+y)2xy, substituting y=vx in the rhs, we notice that (x+y)2xy=1v+v+2, and from y=vx, by differentiating both sides with respect to x we get, dydx=v+xdvdx, in the original equation we get,
Step 2
v+xdvdx=1v+v+2
xdvdx=1v+2
xdvdx=1+2vv
vdv1+2v=dxx
vdv1+2v=dxx
The LHS integral is a simple U-substitution and the RHS is logex.

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