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Question Number 226721 by Spillover last updated on 11/Dec/25
Answered by breniam last updated on 11/Dec/25
x^2 +y^2 +2xyy′=0 t=ln∣x∣ x=e^t y(x)=z(t) (dy/dx)=(dz/dt)×(dt/dx)=(dz/dt)×(1/x) e^(2t) +z^2 +2zz′=0 z^2 +2zz′=−e^(2t) z(z+2z′)=−e^(2t) z=ab z′=a′b+ab′ ab[a(2b′+b)+2a′b]=e^(−2t) 2b′+b=0 ((b′)/b)=−(1/2) ∫((b′)/b)dt=−(1/2)t+C={u=b} ∫(1/u)du=−(1/2)t+C ln∣u∣=−(1/2)t+C b=u=Ae^(−(1/2)t) 2ae^(−t) a′=e^(−2t) 2a′a=e^(−t) 2∫a′adt=−e^(−t) +C_1 ={u=a}= 2∫udu=u^2 =a^2 =−e^(−t) +C_1 a=±(√(C_1 −e^(−t) )) z=ab=e^(−(1/2)t) (√(C_1 −e^(−t) )) z(t)=y(x) t=ln ∣x∣ y=(√((C_1 −(1/x))/x))
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xyy}'=\mathrm{0} \\ $$$${t}=\mathrm{ln}\mid{x}\mid \\ $$$${x}={e}^{{t}} \\ $$$${y}\left({x}\right)={z}\left({t}\right) \\ $$$$\frac{\mathrm{d}{y}}{\mathrm{d}{x}}=\frac{\mathrm{d}{z}}{\mathrm{d}{t}}×\frac{\mathrm{d}{t}}{\mathrm{d}{x}}=\frac{\mathrm{d}{z}}{\mathrm{d}{t}}×\frac{\mathrm{1}}{{x}} \\ $$$${e}^{\mathrm{2}{t}} +{z}^{\mathrm{2}} +\mathrm{2}{zz}'=\mathrm{0} \\ $$$${z}^{\mathrm{2}} +\mathrm{2}{zz}'=−{e}^{\mathrm{2}{t}} \\ $$$${z}\left({z}+\mathrm{2}{z}'\right)=−{e}^{\mathrm{2}{t}} \\ $$$${z}={ab} \\ $$$${z}'={a}'{b}+{ab}' \\ $$$${ab}\left[{a}\left(\mathrm{2}{b}'+{b}\right)+\mathrm{2}{a}'{b}\right]={e}^{−\mathrm{2}{t}} \\ $$$$\mathrm{2}{b}'+{b}=\mathrm{0} \\ $$$$\frac{{b}'}{{b}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int\frac{{b}'}{{b}}\mathrm{d}{t}=−\frac{\mathrm{1}}{\mathrm{2}}{t}+{C}=\left\{{u}={b}\right\} \\ $$$$\int\frac{\mathrm{1}}{{u}}\mathrm{d}{u}=−\frac{\mathrm{1}}{\mathrm{2}}{t}+{C} \\ $$$$\mathrm{ln}\mid{u}\mid=−\frac{\mathrm{1}}{\mathrm{2}}{t}+{C} \\ $$$${b}={u}={Ae}^{−\frac{\mathrm{1}}{\mathrm{2}}{t}} \\ $$$$\mathrm{2}{ae}^{−{t}} {a}'={e}^{−\mathrm{2}{t}} \\ $$$$\mathrm{2}{a}'{a}={e}^{−{t}} \\ $$$$\mathrm{2}\int{a}'{a}\mathrm{d}{t}=−{e}^{−{t}} +{C}_{\mathrm{1}} =\left\{{u}={a}\right\}= \\ $$$$\mathrm{2}\int{u}\mathrm{d}{u}={u}^{\mathrm{2}} ={a}^{\mathrm{2}} =−{e}^{−{t}} +{C}_{\mathrm{1}} \\ $$$${a}=\pm\sqrt{{C}_{\mathrm{1}} −{e}^{−{t}} } \\ $$$${z}={ab}={e}^{−\frac{\mathrm{1}}{\mathrm{2}}{t}} \sqrt{{C}_{\mathrm{1}} −{e}^{−{t}} } \\ $$$${z}\left({t}\right)={y}\left({x}\right) \\ $$$${t}=\mathrm{ln}\:\mid{x}\mid \\ $$$${y}=\sqrt{\frac{{C}_{\mathrm{1}} −\frac{\mathrm{1}}{{x}}}{{x}}} \\ $$
Answered by mr W last updated on 11/Dec/25
x^2 +y^2 +x((d(y^2 ))/dx)=0 let z=y^2 x^2 +z+x(dz/dx)=0 (dz/dx)+(z/x)=−x (type y′+p(x)y=q(x)) ∫(dx/x)=ln x u=e^(∫(dx/x)) =e^(ln x) =x zu=∫x×(−x)dx=−(x^3 /3)+C z=(1/u)(−(x^3 /3)+C)=−(x^2 /3)+(C/x) i.e. y^2 =−(x^2 /3)+(C/x) ✓
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}\frac{{d}\left({y}^{\mathrm{2}} \right)}{{dx}}=\mathrm{0} \\ $$$${let}\:{z}={y}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{z}+{x}\frac{{dz}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dz}}{{dx}}+\frac{{z}}{{x}}=−{x}\:\:\:\:\:\:\left({type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right)\right) \\ $$$$\int\frac{{dx}}{{x}}=\mathrm{ln}\:{x} \\ $$$${u}={e}^{\int\frac{{dx}}{{x}}} ={e}^{\mathrm{ln}\:{x}} ={x} \\ $$$${zu}=\int{x}×\left(−{x}\right){dx}=−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{C} \\ $$$${z}=\frac{\mathrm{1}}{{u}}\left(−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{C}\right)=−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{{C}}{{x}}\: \\ $$$${i}.{e}.\:{y}^{\mathrm{2}} =−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{{C}}{{x}}\:\checkmark \\ $$
Commented by mr W last updated on 11/Dec/25
check: 2yy′=−((2x)/3)−(C/x^2 ) 2xyy′=−((2x^2 )/3)−(C/x) x^2 +y^2 +2xyy′=x^2 −(x^2 /3)+(C/x)−((2x^2 )/3)−(C/x) x^2 +y^2 +2xyy′=0 ✓
$${check}: \\ $$$$\mathrm{2}{yy}'=−\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{{C}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{xyy}'=−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{C}}{{x}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xyy}'={x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{{C}}{{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{C}}{{x}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xyy}'=\mathrm{0}\:\checkmark \\ $$
Answered by Simurdiera last updated on 11/Dec/25
Solucio^� n (x^2 + y^2 ) + 2xy y′ = 0 Cambio de variable y = ux y′ = xu′ + u Reemplazando (x^2 + (ux)^2 ) + 2ux^2 (xu′ + u) = 0 x^2 (1 + u^2 ) + 2ux^2 (xu′ + u) = 0 (1 + u^2 ) + 2uxu′ + 2u^2 = 0 (1 + 3u^2 ) + 2uxu′ = 0 ∫(1/x) dx + ∫((2u)/(1 + 3u^2 )) du = ∫0 ln∣x∣ + (1/3) ln∣1 + 3u^2 ∣ = ln∣K∣ Usando propiedades de logaritmos x∙((1 + 3u^2 ))^(1/3) = K Regresando del cambio de variable ((x∙((x^2 + 3y^2 ))^(1/3) )/( (x^2 )^(1/3) )) = K x^3 + 3xy^2 = C determinant (((y = (√((C − x^3 )/(3x)))))) Resp.
$${Soluci}\acute {{o}n} \\ $$$$\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \right)\:+\:\mathrm{2}{xy}\:{y}'\:=\:\mathrm{0} \\ $$$$\mathrm{Cambio}\:\mathrm{de}\:\mathrm{variable} \\ $$$${y}\:=\:{ux} \\ $$$${y}'\:=\:{xu}'\:+\:{u} \\ $$$$\mathrm{Reemplazando} \\ $$$$\left({x}^{\mathrm{2}} \:+\:\left({ux}\right)^{\mathrm{2}} \right)\:+\:\mathrm{2}{ux}^{\mathrm{2}} \left({xu}'\:+\:{u}\right)\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}\:+\:{u}^{\mathrm{2}} \right)\:+\:\mathrm{2}{ux}^{\mathrm{2}} \left({xu}'\:+\:{u}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\:+\:{u}^{\mathrm{2}} \right)\:+\:\mathrm{2}{uxu}'\:+\:\mathrm{2}{u}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\:+\:\mathrm{3}{u}^{\mathrm{2}} \right)\:+\:\mathrm{2}{uxu}'\:=\:\mathrm{0} \\ $$$$\int\frac{\mathrm{1}}{{x}}\:{dx}\:+\:\int\frac{\mathrm{2}{u}}{\mathrm{1}\:+\:\mathrm{3}{u}^{\mathrm{2}} }\:{du}\:=\:\int\mathrm{0} \\ $$$$\mathrm{ln}\mid{x}\mid\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{ln}\mid\mathrm{1}\:+\:\mathrm{3}{u}^{\mathrm{2}} \mid\:=\:\mathrm{ln}\mid\boldsymbol{{K}}\mid\: \\ $$$$\mathrm{Usando}\:\mathrm{propiedades}\:\mathrm{de}\:\mathrm{logaritmos} \\ $$$${x}\centerdot\sqrt[{\mathrm{3}}]{\mathrm{1}\:+\:\mathrm{3}{u}^{\mathrm{2}} }\:=\:\boldsymbol{{K}} \\ $$$$\mathrm{Regresando}\:\mathrm{del}\:\mathrm{cambio}\:\mathrm{de}\:\mathrm{variable} \\ $$$$\frac{{x}\centerdot\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} \:+\:\mathrm{3}{y}^{\mathrm{2}} }}{\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:=\:\boldsymbol{{K}} \\ $$$${x}^{\mathrm{3}} \:+\:\mathrm{3}{xy}^{\mathrm{2}} \:=\:\boldsymbol{{C}} \\ $$$$\begin{array}{|c|}{{y}\:=\:\sqrt{\frac{\boldsymbol{{C}}\:−\:{x}^{\mathrm{3}} }{\mathrm{3}{x}}}}\\\hline\end{array}\:\:\:{Resp}. \\ $$

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