Question Number 88238 by M±th+et£s last updated on 09/Apr/20
$${solve}\: \\ $$$$\left(\mathrm{3}{x}^{\mathrm{5}} {y}^{\mathrm{4}} +\mathrm{4}{y}\right){dx}+\left(\mathrm{2}{x}^{\mathrm{6}} {y}^{\mathrm{3}} +\mathrm{3}{x}\right){dy}=\mathrm{0} \\ $$
Answered by ajfour last updated on 09/Apr/20
$${dividing}\:{by}\:{x}^{\mathrm{3}} {y}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} {dx}+\mathrm{2}{x}^{\mathrm{3}} {ydy}\right)+\frac{\left(\mathrm{4}{ydx}+\mathrm{3}{xdy}\right)}{{x}^{\mathrm{3}} {y}^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{3}} {y}^{\mathrm{2}} {d}\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)+\left(\mathrm{4}{ydx}+\mathrm{3}{xdy}\right)=\mathrm{0} \\ $$$${multiplying}\:{by}\:{x}^{\mathrm{3}} {y}^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)^{\mathrm{2}} {d}\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)+\left(\mathrm{4}{x}^{\mathrm{3}} {y}^{\mathrm{3}} {dx}+\mathrm{3}{x}^{\mathrm{4}} {y}^{\mathrm{2}} {dy}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)^{\mathrm{2}} {d}\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)+{d}\left({x}^{\mathrm{4}} {y}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\frac{\left({x}^{\mathrm{3}} {y}^{\mathrm{2}} \right)^{\mathrm{3}} }{\mathrm{3}}+{x}^{\mathrm{4}} {y}^{\mathrm{3}} =\:{c}\:. \\ $$
Commented by M±th+et£s last updated on 09/Apr/20
$${nice}\:{solution}\:.\:{hod}\:{bless}\:{you} \\ $$