Question Number 116053 by Study last updated on 30/Sep/20
$${y}\frac{{dy}}{{dx}}=\mathrm{1}+{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:{y}=? \\ $$
Answered by Dwaipayan Shikari last updated on 30/Sep/20
$$\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{1}+\mathrm{x}^{\mathrm{2}} \\ $$$$\int\mathrm{ydy}=\int\mathrm{1}+\mathrm{x}^{\mathrm{2}} \mathrm{dx} \\ $$$$\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{C} \\ $$$$\mathrm{y}=\pm\sqrt{\mathrm{2x}+\frac{\mathrm{2x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2C}} \\ $$
Answered by Bird last updated on 30/Sep/20
$${yy}^{'} \:=\mathrm{1}+{x}^{\mathrm{2}\:} \:\Rightarrow\int{yy}^{'} {dx}\:={x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+{c} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} \:={x}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{c}\:\Rightarrow \\ $$$${y}^{\mathrm{2}\:} =\mathrm{2}{x}+\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}\:+\mathrm{2}{c}\:\Rightarrow \\ $$$${y}\:=\overset{−} {+}\sqrt{\mathrm{2}{x}+\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{3}}+\lambda}\:\:\:\left(\lambda=\mathrm{2}{c}\right) \\ $$