wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve the differential equation yexydx=xexy+y2dy(y0).

Open in App
Solution


The differential equation is
yexydx=(xexy+y2)dy
let's take p as p=exy
so differential equation will become
ypdx=(xp+y2)dy
differentiating p with respect to x

dp=exy(dxyxdyy2)

y2dpdx=p(yxdydx)

pydx=y2dp+pxdy
apply this on above differential equation

xpdy+y2dp=xpdy+y2dy

dy=dp,

y=p+C,
hence the solution is :
y=exy+C

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon