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

Show that the general solution of the differential equation dydx+y2+y+1x2+x+1=0 is given by (x+y+1)=A(1xy2xy), where A is a parameter.

Open in App
Solution

Given, differential equation is
dydx+y2+y+1x2+x+1=0dyy2+y+1+dxx2+x+1=0
On integrating both sides, we get
dyy2+y+1+dxx2+x+1=Cdyy2+y+1+(12)2(12)2+dxx2+x+1+(12)2(12)2=Cdy(y+12)2+(114)+dx(x+12)2+(114)=Cdy(y+12)2+(32)2+dx(x+12)2+(114)=C23tan1(y+1232)+23tan1(x+1232)=C
[1a2+x2dx=1atan1xa]
tan1(2y+13)+tan1(2x+13)=3C2=k (say)
tan12y+13+2x+131(2y+13)(2x+13)=k
[tan1x+tan1y=tan1(x+y1xy)]
tan12y+1+2x+131(4xy+2x+2y+13)=k23(x+y+1)3(4xy+2x+2y+1)=tan k23(x+y+1)2(1xy2xy)=tan kx+y+1=13tan k(1xy2xy)
Let, A=13tan k is an arbitrary constant.
x+y+1=A(1xy2xy)
Hence, the given result is proved.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebra of Derivatives
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon