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Question

Find the particular solution of the differential equation2y exydx+(y2x exy)dy=0 given that x=0 when y=1.

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Solution

We have

2y exydx+(y2x exy)dy=0

dxdy=2x exyy2y exy ...............(1)

Here, the given differential equation is a homogeneous differential equation. As the right hand side of equation (1) is expressed as function of xy.

Therefore, we put x=vy

dxdy=v+ydvdy

substituting x=vy in the given differential equation we will get,

ydvdy=2vev12evv

ydvdy=12ev

2ev dv=1ydy,y0

2ev dv=1ydy

2ev=log|y|+C

2exy+log|y|=C ...........(2)

It is given that x=0 when y=1,

So putting x=0,y=1 in equation (2), we get

2e0+log 1=C

C=2

Putting C=2 in equation (2), we get the required particular solution

2 exy+log y=2

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