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

If y=x5(cos(lnx)+sin(lnx)), then the value of (a+b) in the relation x2y2+axy1+by=0 is
(y1 and y2 denote the first and second derivative of y with respect to x respectively.)

A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
17
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
23
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 17
We have y=x5(cos(lnx)+sin(lnx))
dydx=5x4(cos(lnx)+sin(lnx))+x5(sin(lnx)x+cos(lnx)x)
xy1=5y+x5(cos(lnx)sin(lnx))
Again differentiating w.r.t. x,
xy2+y1=5y1+5x4(cos(lnx)sin(lnx))+x5(sin(lnx)xcos(lnx)x)
x2y2+xy1=5xy1+5x5(cos(lnx)sin(lnx))x5(sin(lnx)+cos(lnx))
x2y24xy1=5(xy15y)y
x2y24xy1=5xy126y
x2y29xy1+26y=0
Comparing with x2y2+axy1+by=0, we get
a=9 and b=26
Hence, a+b=17

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