Question

If
y = Sin 'x
show that (1-x²) d^2y/dx^2 + x d y/dx = 0​

Answer 1

Step-by-step explanation:

I take d^2y/dx is y2

and similarily dy/dx as y

so y= sin'x

y1 =1/√(1-x^2)

squaring both side

y1^2= 1/(1-x^2)

(1-x^2)y1=1

again diff. both side

(1-x^2)y2-2xy1=0(by application of chain rule)

hence proved....

hope it will help you dear....