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`ysin2xdx-(1+y^2+cos^2x)dy=0` |
Answer» `Pdx+Qdy=0` `ysin2xdx+(-1-y^2-cos^2x)dy=0` `(dp)/(dy)=sin2x` `(dQ)/(dx)=-2cosx*(-sinx)=2sinxcosx` `intPdx=-1/2ycos2x+g(y)` `intQdy=-y-y^33/3-ycos^2x+h(x)` `cos^2x=1/2(1+cos2x)` `g(y):-y-y^3/3-y/2` and `h(x)=0` `-1/2ycos2x-3/2y-y^3/3=0`. |
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