CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For each the differential equations given, find the general solution :
(1+x2)dy+2xydx=cotxdx(x0)

Open in App
Solution

(1+x2)dy+2xydx=cotxdx

dydx+(2x1+x2)y=cotx1+x2

dydx+Py=Q

P=2x1+x2,Q=cotx1+x2

I.F.=ePdx

=e2x1+x2dx

=1+x2

y×I.F=Q×I.Fdx+c

y×(1+x2)=(cotx1+x2)(1+x2)dx+c

y×(1+x2)=cotxdx+c

y(1+x2)=logsinx+c

y=(1+x2)1logsinx+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