We have,
y=sin(m(sin−1x)) ……… (1)
On differentiating both sides w.r.t x, we get
dydx=cos(m(sin−1x))×m√1−x2
dydx=mcos(m(sin−1x))√1−x2
On differentiating both sides w.r.t x, we get
d2ydx2=m⎡⎢ ⎢ ⎢ ⎢⎣(√1−x2)(−sin(m(sin−1x)))×m√1−x2−(cos(m(sin−1x)))×12√1−x2×(0−2x)(√1−x2)2⎤⎥ ⎥ ⎥ ⎥⎦
d2ydx2=m⎡⎢ ⎢ ⎢⎣(−msin(m(sin−1x)))+(cos(m(sin−1x)))×x√1−x21−x2⎤⎥ ⎥ ⎥⎦
d2ydx2=m⎡⎢ ⎢⎣−msin(m(sin−1x))1−x2+xcos(m(sin−1x))(1−x2)(√1−x2)⎤⎥ ⎥⎦
d2ydx2=m⎡⎢ ⎢ ⎢⎣−msin(m(sin−1x))1−x2+xdydx(1−x2)⎤⎥ ⎥ ⎥⎦....[dydx=mcos(m(sin−1x))√1−x2]
(1−x2)d2ydx2=mxdydx−m2sin(m(sin−1x))
(1−x2)d2ydx2=mxdydx−m2y.......[y=sin(m(sin−1x))]
(1−x2)d2ydx2−mxdydx+m2y=0
Hence, the value is 0.