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Question

The curve satisfying dydx=siny+xsin2yxcosy passes through (1,π) and (h,π2) then the maximum value of |h| is

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Solution

dydx=siny+xsin2yxcosycosydydx=siny+x2sinyx

Let siny=tdt=cosy dy
dtdx=t+x2tx
Assuming t=vxdtdx=v+xdvdx
v+xdvdx=v+12v1xdvdx=2v2+2v+12v12v12v2+2v+1 dv=dxx

Putting 2v2+2v+1=zdz=(4v+2)dv
12ln|z|=ln|x|+ln|C|12ln2v2+2v+1=ln|x|+ln|C|12ln2sin2yx2+2sinyx+1=ln|x|+ln|C|ln2sin2yx2+2sinyx+1=2ln|Cx|2sin2yx2+2sinyx+1=1(Cx)2sin2y=xsiny+x222C2

The curve passes through (1,π)
0=0+122C2C2=4
Therefore the required curve is,
sin2y=xsiny+x2212
Putting x=h,y=π2
1=h+h2212h2+2h=3(h+1)2=2h=1,3

Hence, the maximum value of |h|=3


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