Question

If log y = m sin 1x, then the correct relation in y, yı, y, is:
(a) (1 - x2)yz - xy. - m²y = 0
(b) (1 – x?)y2 + xyz + m²y = 0
(C) x?y2 – xyı + m2y = 0
(d) (1 + x²)y2 - xyı + y = 0​

Answer 1

Hope this'll help you.

Answer 2

Step-by-step explanation:

:  y′=−sin(msin−1x)1−x2m

y′′=−cos(msin−1x)1−x2m2−sin(msin−1x)(1−x2)3/2mx

y′′=−1−x2ym2+1−x2xy′

(1−x2)y′′+ym2−xy′=0

ii:  yy′=1+x