Question Number 227173 by Spillover last updated on 04/Jan/26
$$\boldsymbol{{If}}\:\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{x}}\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\boldsymbol{{x}}\left(\boldsymbol{{x}}+\mathrm{1}\right)\boldsymbol{{y}}^{\mathrm{3}} \\ $$$${Prove}\:{that}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}/\mathrm{1}/\mathrm{2026} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{6}{x}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +{A}} \\ $$
Answered by Simurdiera last updated on 04/Jan/26
![Solucio^ n y′ − (1/(2x)) y = − ((x + 1)/2) y^3 Ecuacio^ n diferencial de Bernoulli u = y^(1−3) = y^(−2) u′ = −2y^(−3) y′ Reemplazando en la ecuacio^ n diferencial −2y^(−3) y′ − ((−2y^(−3) )/(2x)) y = − ((x + 1)/2)(−2y^(−3) )y^3 −2y^(−3) y′ + (y^(−2) /x) = x + 1 u′ + (1/x) u = x + 1 Ecuacio^ n diferencial Lineal de primer orden u = e^(−∫(1/x) dx) [∫e^(∫(1/x) dx) (x + 1)dx + C] u = e^(−ln(x)) [∫e^(ln(x)) (x + 1)dx + C] u = e^(ln((1/x))) [∫e^(ln(x)) (x + 1)dx + C] Aplicando leyes de exponentes u = (1/x)[∫x(x + 1)dx + C] u = (1/x)[∫(x^2 + x) dx + C] u = (1/x)[(x^3 /3) + (x^2 /2) + C] u = ((2x^3 + 3x^2 + 6C)/(6x)) Regresando del cambio de variable y^2 = ((6x)/(2x^3 + 3x^2 + A)) ✓](https://www.tinkutara.com/question/Q227190.png)
$${Soluci}\acute {{o}n} \\ $$$${y}'\:−\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:{y}\:=\:−\:\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\:{y}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\boldsymbol{{Ecuaci}}\acute {\boldsymbol{{o}n}}\:\boldsymbol{{diferencial}}\:\boldsymbol{{de}}\:\boldsymbol{{Bernoulli}} \\ $$$${u}\:=\:{y}^{\mathrm{1}−\mathrm{3}} \:=\:{y}^{−\mathrm{2}} \\ $$$${u}'\:=\:−\mathrm{2}{y}^{−\mathrm{3}} \:{y}' \\ $$$$\mathrm{Reemplazando}\:\mathrm{en}\:\mathrm{la}\:\mathrm{ecuaci}\acute {\mathrm{o}n}\:\mathrm{diferencial} \\ $$$$−\mathrm{2}{y}^{−\mathrm{3}} {y}'\:−\:\frac{−\mathrm{2}{y}^{−\mathrm{3}} }{\mathrm{2}{x}}\:{y}\:=\:−\:\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}{y}^{−\mathrm{3}} \right){y}^{\mathrm{3}} \\ $$$$−\mathrm{2}{y}^{−\mathrm{3}} {y}'\:+\:\frac{{y}^{−\mathrm{2}} }{{x}}\:=\:{x}\:+\:\mathrm{1} \\ $$$${u}'\:+\:\frac{\mathrm{1}}{{x}}\:{u}\:=\:{x}\:+\:\mathrm{1}\:\:\:\:\:\:\:\:\:\boldsymbol{{Ecuaci}}\acute {\boldsymbol{{o}n}}\:\boldsymbol{{diferencial}}\:\boldsymbol{{Lineal}}\:\boldsymbol{{de}}\:\boldsymbol{{primer}}\:\boldsymbol{{orden}} \\ $$$${u}\:=\:{e}^{−\int\frac{\mathrm{1}}{{x}}\:{dx}} \left[\int{e}^{\int\frac{\mathrm{1}}{{x}}\:{dx}} \left({x}\:+\:\mathrm{1}\right){dx}\:+\:\boldsymbol{{C}}\right] \\ $$$${u}\:=\:{e}^{−\mathrm{ln}\left({x}\right)} \left[\int{e}^{\mathrm{ln}\left({x}\right)} \left({x}\:+\:\mathrm{1}\right){dx}\:+\:\boldsymbol{{C}}\right] \\ $$$${u}\:=\:{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)} \left[\int{e}^{\mathrm{ln}\left({x}\right)} \left({x}\:+\:\mathrm{1}\right){dx}\:+\:\boldsymbol{{C}}\right] \\ $$$$\mathrm{Aplicando}\:\mathrm{leyes}\:\mathrm{de}\:\mathrm{exponentes} \\ $$$${u}\:=\:\frac{\mathrm{1}}{{x}}\left[\int{x}\left({x}\:+\:\mathrm{1}\right){dx}\:+\:\boldsymbol{{C}}\right] \\ $$$${u}\:=\:\frac{\mathrm{1}}{{x}}\left[\int\left({x}^{\mathrm{2}} \:+\:{x}\right)\:{dx}\:+\:\boldsymbol{{C}}\right] \\ $$$${u}\:=\:\frac{\mathrm{1}}{{x}}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\boldsymbol{{C}}\right] \\ $$$${u}\:=\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{6}\boldsymbol{{C}}}{\mathrm{6}{x}} \\ $$$$\mathrm{Regresando}\:\mathrm{del}\:\mathrm{cambio}\:\mathrm{de}\:\mathrm{variable} \\ $$$${y}^{\mathrm{2}} \:=\:\frac{\mathrm{6}{x}}{\mathrm{2}{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\boldsymbol{{A}}}\:\:\:\:\:\checkmark \\ $$
Commented by Spillover last updated on 06/Jan/26
$${thanks} \\ $$