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

Ans fast

Q14. If x = a, cos θ + b sin ​θ and y = a sin θ – b cos θ, show that y2d2ydx2-xdydx+y =0.

Open in App
Solution

Dear Student
Givenx=acosθ+bsinθy=asinθ-bcosθNowx2=a2cos2θ+b2sin2θ+2absinθcosθy2=a2sin2θ+b2cos2θ-2absinθcosθAdding x2+y2 we getx2+y2=a2cos2θ+b2sin2θ+2absinθcosθ+a2sin2θ+b2cos2θ-2absinθcosθ =a2cos2θ+sin2θ+b2cos2θ+sin2θ =a2+b2Now differentiating both sides we get2x+2ydydx=0dydx=-xyagain differentiating we getd2ydx2=-y-xdydxy2y2d2ydx2=-y+xdydxy2d2ydx2-xdydx+y=0Hence proved

Regards

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