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Question Number 96117 by bobhans last updated on 30/May/20

$$(1-2xy)dx+(4y^3-x^2)dy=0$$

Answered by john santu last updated on 30/May/20

Commented by john santu last updated on 30/May/20

$$\mathrm{typo}\mathrm{in}{6}^{\mathrm{th}}\mathrm{line}.$$$$\mathrm{it}\mathrm{should}\mathrm{be}{\mathrm{g}}^{\prime }\left(\mathrm{y}\right)={4\mathrm{y}}^3-{\mathrm{x}}^2-\frac{\partial }{\partial \mathrm{y}}(\int \mathrm{M}(x,\mathrm{y})dx)$$

Commented by bobhans last updated on 30/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by i jagooll last updated on 30/May/20

yeahhh

Commented by Tinku Tara last updated on 02/Jun/20

$$\mathrm{Hi}\mathrm{jagooll}$$$$\mathrm{Are}\mathrm{you}\mathrm{able}\mathrm{to}\mathrm{post}\mathrm{questions}$$$$\mathrm{now}.$$

Answered by bemath last updated on 30/May/20

$$inotherway$$$$dx-(2xydx+x^{2}dy)+4y^{3}dy=0$$$$dx-d\left(x^{2}y\right)+4y^{3}dy=0$$$$\int dx-\int d\left(x^{2}y\right)+\int 4y^{3}dy=C$$$$x-x^{2}y+y^4=C$$

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