x2ydx+(y+x3)dy=0 ⟹ x2ydx=−(y+x3)dydxdy=−y−x3x2ydxdy=−1x2−xy ⟹ dxdy+xy=−1x2dxdy+Rx=SR=1y, S=−1x2IF=ϵ∫Rdy=ϵ∫1ydy=ϵlog y=yx⋅IF=∫S⋅IF dy+Cx⋅y=∫−1x2⋅y dy+C=∫−yx2 dy+C=−y22x2+C∴xy=−y22x2+C−−−−−−−−>Answerx^2 ydx + (y +x^3)dy=0 \implies x^2ydx=-(y +x^3)dy\\ \frac{dx}{dy}=\frac{-y-x^3}{x^2y}\\ \frac{dx}{dy}=-\frac{1}{x^2}-\frac{x}{y}\implies \frac{dx}{dy}+ \frac{x}{y}= -\frac{1}{x^2}\\ \frac{dx}{dy}+ Rx= S\\ R=\frac{1}{y},\ S=-\frac{1}{x^2}\\ IF=\epsilon^{\int Rdy}=\epsilon^{\int \frac{1}{y}dy}=\epsilon^{log\ y}=y\\ x \cdot IF = \int S \cdot IF\ dy+C\\ x \cdot y = \int -\frac{1}{x^2} \cdot y\ dy+C= \int -\frac{y}{x^2}\ dy+C= -\frac{y^2}{2x^2}+C\\ \therefore xy=-\frac{y^2}{2x^2}+C-------->Answerx2ydx+(y+x3)dy=0⟹x2ydx=−(y+x3)dydydx=x2y−y−x3dydx=−x21−yx⟹dydx+yx=−x21dydx+Rx=SR=y1, S=−x21IF=ϵ∫Rdy=ϵ∫y1dy=ϵlog y=yx⋅IF=∫S⋅IF dy+Cx⋅y=∫−x21⋅y dy+C=∫−x2y dy+C=−2x2y2+C∴xy=−2x2y2+C−−−−−−−−>Answer
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