Question Number 169389 by Shrinava last updated on 29/Apr/22
$$\mathrm{1}.\:\mathrm{2}\sqrt{\mathrm{y}}\:\mathrm{dx}\:=\:\mathrm{dy} \\ $$$$\mathrm{2}.\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2xydy} \\ $$$$\mathrm{3}.\:\mathrm{xdx}\:+\:\frac{\mathrm{1}}{\mathrm{y}}\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{4}.\:\mathrm{dy}\:=\:\mathrm{3x}^{\mathrm{2}} \:\mathrm{dx} \\ $$$$\mathrm{5}.\:\mathrm{2y}^{\mathrm{2}} \mathrm{dx}\:+\:\mathrm{x}\left(\mathrm{1}\:+\:\mathrm{y}^{\mathrm{2}} \right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$$$\mathrm{6}.\:\mathrm{4x}^{\mathrm{3}} \mathrm{dx}\:+\:\mathrm{dy}\:=\:\mathrm{0} \\ $$
Commented by mkam last updated on 29/Apr/22
$$\left(\mathrm{1}\right)\:{dx}\:=\:\frac{{dy}}{\mathrm{2}\sqrt{{y}}}\:\:\Rightarrow\:{x}\:=\:\sqrt{{y}}\:+\:{C}\:\Rightarrow\:{y}\:=\:\left(\:{x}+{c}\:\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\frac{{dy}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}\:\Rightarrow\:{v}\:+\:{x}\:\frac{{dv}}{{dx}}\:=\:\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{v}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} \:{v}}\:\Rightarrow\:\:{x}\frac{{dv}}{{dx}}\:=\:\frac{\mathrm{1}+{v}^{\mathrm{2}} }{\mathrm{2}{v}}\:−\:{v} \\ $$$$ \\ $$$$\Rightarrow\:{x}\:\frac{{dv}}{{dx}\:}=\:\frac{\mathrm{1}−{v}^{\mathrm{2}} }{\mathrm{2}{v}}\:\Rightarrow\:\frac{{dx}}{{x}}\:=\:\frac{\mathrm{2}{v}}{−\left({v}^{\mathrm{2}} −\mathrm{1}\right)}{dv}\:\Rightarrow\:{lnx}\:+\:{lnc}\:=\:−\:\mathrm{2}{ln}\mid{v}^{\mathrm{2}} −\mathrm{1}\mid \\ $$$$ \\ $$$$\Rightarrow\:{ln}\left({xc}\right)=\:{ln}\mid\frac{\mathrm{1}}{\left({v}^{\mathrm{2}} −\mathrm{1}\right)}\mid\:\Rightarrow\:{xc}\:=\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{3}\right){xdx}+\frac{\mathrm{1}}{{y}}{dy}=\mathrm{0}\:\Rightarrow\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:{lny}\:=\:{C} \\ $$$$ \\ $$$$\left(\mathrm{4}\right){dy}\:=\:\mathrm{3}{x}^{\mathrm{2}} {dx}\:\Rightarrow\:{y}\:=\:{x}^{\mathrm{3}} +\:{C} \\ $$$$ \\ $$$$\left(\mathrm{5}\right)\mathrm{2}{y}^{\mathrm{2}} {dx}+{x}\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0}\:\Rightarrow\:\frac{\mathrm{2}{dx}}{{x}}\:+\:\left(\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\:+\:\mathrm{1}\:\right){dy}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\:\mathrm{2}{lnx}\:−\frac{\mathrm{1}}{{y}}\:+\:{y}\:=\:{C} \\ $$$$ \\ $$$$\left(\mathrm{6}\right)\mathrm{4}{x}^{\mathrm{3}} {dx}\:+\:{dy}\:=\:\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{4}} +{y}\:=\:{C} \\ $$$$ \\ $$$${aldolaimy} \\ $$$$ \\ $$