Question Number 92708 by Rio Michael last updated on 08/May/20
$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}. \\ $$$$\:\left(\mathrm{a}\right)\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{b}\right)\:\left(\mathrm{2}{y}−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:−\mathrm{2}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$
Answered by peter frank last updated on 09/May/20
$${let}\:{p}={y}^{'} \\ $$$$\frac{{dp}}{{dx}}={y}'' \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\frac{{dp}}{{dx}}=\frac{{dp}}{{dy}}.\frac{{dy}}{{dx}}=\frac{{dp}}{{dy}}.{p} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}.{p}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$…….. \\ $$$$ \\ $$$$ \\ $$
Commented by Rio Michael last updated on 10/May/20
$$\mathrm{okay}\:\mathrm{bro}\:\mathrm{thats}\:\mathrm{great} \\ $$