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Question Number 92708 by Rio Michael last updated on 08/May/20

solve the differential equations.   (a) (x + 3y^2 )(d^2 y/dx^2 ) + 6y ((dy/dx))^2  + 2(dy/dx) + 2 = 0  (b) (2y−x)(d^2 y/dx^2 ) + 2((dy/dx))^2  −2 (dy/dx) + 2 = 0

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}. \\ $$$$\:\left(\mathrm{a}\right)\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{b}\right)\:\left(\mathrm{2}{y}−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:−\mathrm{2}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$

Answered by peter frank last updated on 09/May/20

let p=y^′   (dp/dx)=y′′    (x + 3y^2 )(d^2 y/dx^2 ) + 6y ((dy/dx))^2  + 2(dy/dx) + 2 = 0    (x + 3y^2 )(dp/dx) + 6y (p)^2  + 2p + 2 = 0    (x + 3y^2 )(dp/dx) + 6y (p)^2  + 2p + 2 = 0  (dp/dx)=(dp/dy).(dy/dx)=(dp/dy).p    (x + 3y^2 )(dp/dx).p + 6y (p)^2  + 2p + 2 = 0  ........

$${let}\:{p}={y}^{'} \\ $$$$\frac{{dp}}{{dx}}={y}'' \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\frac{{dp}}{{dx}}=\frac{{dp}}{{dy}}.\frac{{dy}}{{dx}}=\frac{{dp}}{{dy}}.{p} \\ $$$$\:\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{dp}}{{dx}}.{p}\:+\:\mathrm{6}{y}\:\left({p}\right)^{\mathrm{2}} \:+\:\mathrm{2}{p}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$........ \\ $$$$ \\ $$$$ \\ $$

Commented by Rio Michael last updated on 10/May/20

okay bro thats great

$$\mathrm{okay}\:\mathrm{bro}\:\mathrm{thats}\:\mathrm{great} \\ $$

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