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Question

If y=sin(msin1x), then (1x2)d2ydx2xdydx+m2y=

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Solution

We have,

y=sin(m(sin1x)) ……… (1)

On differentiating both sides w.r.t x, we get

dydx=cos(m(sin1x))×m1x2

dydx=mcos(m(sin1x))1x2

On differentiating both sides w.r.t x, we get

d2ydx2=m⎢ ⎢ ⎢ ⎢(1x2)(sin(m(sin1x)))×m1x2(cos(m(sin1x)))×121x2×(02x)(1x2)2⎥ ⎥ ⎥ ⎥

d2ydx2=m⎢ ⎢ ⎢(msin(m(sin1x)))+(cos(m(sin1x)))×x1x21x2⎥ ⎥ ⎥

d2ydx2=m⎢ ⎢msin(m(sin1x))1x2+xcos(m(sin1x))(1x2)(1x2)⎥ ⎥

d2ydx2=m⎢ ⎢ ⎢msin(m(sin1x))1x2+xdydx(1x2)⎥ ⎥ ⎥....[dydx=mcos(m(sin1x))1x2]

(1x2)d2ydx2=mxdydxm2sin(m(sin1x))

(1x2)d2ydx2=mxdydxm2y.......[y=sin(m(sin1x))]

(1x2)d2ydx2mxdydx+m2y=0

Hence, the value is 0.



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