dydx=siny+xsin2y−xcosy⇒cosydydx=siny+x2siny−x
Let siny=t⇒dt=cosy dy
⇒dtdx=t+x2t−x
Assuming t=vx⇒dtdx=v+xdvdx
⇒v+xdvdx=v+12v−1⇒xdvdx=−2v2+2v+12v−1⇒∫2v−1−2v2+2v+1 dv=∫dxx
Putting −2v2+2v+1=z⇒dz=(−4v+2)dv
⇒−12ln|z|=ln|x|+ln|C|⇒−12ln∣∣−2v2+2v+1∣∣=ln|x|+ln|C|⇒−12ln∣∣∣−2sin2yx2+2sinyx+1∣∣∣=ln|x|+ln|C|⇒ln∣∣∣−2sin2yx2+2sinyx+1∣∣∣=−2ln|Cx|⇒−2sin2yx2+2sinyx+1=1(Cx)2⇒sin2y=xsiny+x22−2C2
The curve passes through (1,π)
0=0+12−2C2⇒C2=4
Therefore the required curve is,
sin2y=xsiny+x22−12
Putting x=h,y=π2
1=h+h22−12⇒h2+2h=3⇒(h+1)2=2⇒h=1,−3
Hence, the maximum value of |h|=3