Let u=sinx . We have cosxdx=du and (z+z3)du−(z+z3)dy+(1−z2)(y−u)dz=0 . That is (z+z3)(du−dy)+(1−z2)(y−u)dz=0 .
Next, let u−y=v . Since dv=du−dy , we obtain (z+z3)dv+(1−z2)(−v)dz=0.
Separate variables: vdv=z+z31−z2dz We know that vdv=dln∣v∣ . Consider z+z31−z2 .z+z3=z(z2+1) , so a partial fraction decomposition of z+z31−z2 is zA+z2+1Bz+C==z(z2+1)A(z2+1)+z(Bz+C) and therefore 1−z2=A(z2+1)+z(Bz+C)=Az2+A+Bz2+Cz .
Equality of the polynomials 1−z2 and Az2+A+Bz2+Cz=(A+B)z2+Cz+A gives us a system of equations
⎩⎨⎧A+B=−1C=0A=1 So, A=1,B=−2,C=0 , and z+z31−z2=z1−z2+12z . We have z+z31−z2dz=zdz−z2+12zdz=
=dln∣z∣−z2+1dz2=dln∣z∣−z2+1d(z2+1)=
=dln∣z∣−dln∣z2+1∣=dln∣∣z2+1z∣∣ .
Since dln∣v∣=dln∣∣z2+1z∣∣ , we have ln∣v∣=ln∣∣z2+1z∣∣+D . Take the exponent of the equation: ∣v∣=eD∣∣z2+1z∣∣ . That is, v=eDz2+1z or v=−eDz2+1z . The case v=0 also is a solution of the equation (z+z3)dv+(1−z2)(−v)dz=0.
Union of the cases v=eDz2+1z, v=−eDz2+1z and v=0 is the case v=Ez2+1z .
Remembering that v=u−y and u=sinx , we have sinx−y=Ez2+1z
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