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Question Number 5097 by LMTV last updated on 12/Apr/16
x^2 y′=x^2 +xy+y^2 yy′+sin x=0 y′=1−y^2 separation of variables
$${x}^{\mathrm{2}} {y}'={x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \\ $$$${yy}'+\mathrm{sin}\:{x}=\mathrm{0} \\ $$$${y}'=\mathrm{1}−{y}^{\mathrm{2}} \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{variables} \\ $$
Answered by Yozzii last updated on 12/Apr/16
y^′ =((x(x+y))/(x^2 −1)) y^′ −(x/(x^2 −1))y=(x^2 /(x^2 −1)) y^′ +P(x)y=Q(x) I(x)=e^(−∫(x/(x^2 −1))dx) =e^(−(1/2)∫((2x)/(x^2 −1))dx) =e^(−0.5ln(x^2 −1)) I(x)=(1/( (√(x^2 −1)))) Using yI(x)=∫I(x)Q(x)dx yI(x)=∫(x^2 /((x^2 −1)^(3/2) ))dx x=coshu⇒dx=sinhudu x^2 −1=cosh^2 u−1=sinh^2 u. yI(x)=∫((cosh^2 usinhu)/(sinh^3 u))du yI(x)=∫((cosh^2 u)/(sinh^2 u))du=∫(cosech^2 u+1)du D(cothu)=((sinhusinhu−coshucoshu)/(sinh^2 u)) D(cothu)=((−(cosh^2 u−sinh^2 u))/(sinh^2 u)) D(cothu)=−cosech^2 u ⇒cothu=−∫cosech^2 udu yI(x)=−cothu+u+C u=cosh^(−1) x ⇒y=(C+cosh^(−1) x−coth(cosh^(−1) x))(√(x^2 −1)) y=(C+ln(x+(√(x^2 −1)))−((cosh(cosh^(−1) x))/(sinh(cosh^(−1) x))))(√(x^2 −1)) sinh(cosh^(−1) x)=((e^(ln(x+(√(x^2 −1)))) −e^(ln(1/(x+(√(x^2 −1))))) )/2) sinh(cosh^(−1) x)=(((x+(√(x^2 −1)))^2 −1)/(2(x+(√(x^2 −1))))) cosh(cosh^(−1) x)=(((x+(√(x^2 −1)))^2 +1)/(2(x+(√(x^2 −1))))) ⇒coth(cosh^(−1) x)=(((x+(√(x^2 −1)))^2 +1)/((x+(√(x^2 −1)))^2 −1)) coth(cosh^(−1) x)=((x^2 +2x(√(x^2 −1))+x^2 −1+1)/(x^2 +2x(√(x^2 −1))+x^2 −1−1)) coth(cosh^(−1) x)=((x(x+(√(x^2 −1))))/(x^2 +x(√(x^2 −1))−1)) =(x/(x−(((x−(√(x^2 −1))))/(x^2 −x^2 +1)))) =(x/(x−x+(√(x^2 −1)))) coth(cosh^(−1) x)=(x/( (√(x^2 −1)))) ∴ y=(C+ln(x+(√(x^2 −1))))−x
$${y}^{'} =\frac{{x}\left({x}+{y}\right)}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${y}^{'} −\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{y}=\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${y}^{'} +{P}\left({x}\right){y}={Q}\left({x}\right) \\ $$$${I}\left({x}\right)={e}^{−\int\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}} ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}{dx}} ={e}^{−\mathrm{0}.\mathrm{5}{ln}\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${I}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${Using}\:{yI}\left({x}\right)=\int{I}\left({x}\right){Q}\left({x}\right){dx} \\ $$$${yI}\left({x}\right)=\int\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dx} \\ $$$${x}={coshu}\Rightarrow{dx}={sinhudu} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}={cosh}^{\mathrm{2}} {u}−\mathrm{1}={sinh}^{\mathrm{2}} {u}. \\ $$$${yI}\left({x}\right)=\int\frac{{cosh}^{\mathrm{2}} {usinhu}}{{sinh}^{\mathrm{3}} {u}}{du} \\ $$$${yI}\left({x}\right)=\int\frac{{cosh}^{\mathrm{2}} {u}}{{sinh}^{\mathrm{2}} {u}}{du}=\int\left({cosech}^{\mathrm{2}} {u}+\mathrm{1}\right){du} \\ $$$${D}\left({cothu}\right)=\frac{{sinhusinhu}−{coshucoshu}}{{sinh}^{\mathrm{2}} {u}} \\ $$$${D}\left({cothu}\right)=\frac{−\left({cosh}^{\mathrm{2}} {u}−{sinh}^{\mathrm{2}} {u}\right)}{{sinh}^{\mathrm{2}} {u}} \\ $$$${D}\left({cothu}\right)=−{cosech}^{\mathrm{2}} {u} \\ $$$$\Rightarrow{cothu}=−\int{cosech}^{\mathrm{2}} {udu} \\ $$$${yI}\left({x}\right)=−{cothu}+{u}+{C} \\ $$$${u}={cosh}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow{y}=\left({C}+{cosh}^{−\mathrm{1}} {x}−{coth}\left({cosh}^{−\mathrm{1}} {x}\right)\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${y}=\left({C}+{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)−\frac{{cosh}\left({cosh}^{−\mathrm{1}} {x}\right)}{{sinh}\left({cosh}^{−\mathrm{1}} {x}\right)}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${sinh}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{{e}^{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} −{e}^{{ln}\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}} }{\mathrm{2}} \\ $$$${sinh}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$${cosh}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$\Rightarrow{coth}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}}{\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$${coth}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+{x}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{1}} \\ $$$${coth}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{{x}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{{x}^{\mathrm{2}} +{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}} \\ $$$$=\frac{{x}}{{x}−\frac{\left({x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{{x}^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\frac{{x}}{{x}−{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${coth}\left({cosh}^{−\mathrm{1}} {x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\therefore\:{y}=\left({C}+{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right)−{x} \\ $$$$ \\ $$$$ \\ $$
Commented by LMTV last updated on 12/Apr/16
sorry ToT change question... my miss...
$$\mathrm{sorry}\:\mathrm{ToT}\:\mathrm{change}\:\mathrm{question}…\:\mathrm{my}\:\mathrm{miss}… \\ $$
Commented by Yozzii last updated on 12/Apr/16
Answer based on question posted before correction: x^2 y^′ =x^2 +xy+y^′ .
$${Answer}\:{based}\:{on}\:{question}\:{posted}\:{before} \\ $$$${correction}:\:{x}^{\mathrm{2}} {y}^{'} ={x}^{\mathrm{2}} +{xy}+{y}^{'} . \\ $$
Commented by Yozzii last updated on 12/Apr/16
Pas de proble^� me.
$${Pas}\:{de}\:{probl}\grave {{e}me}. \\ $$
Answered by Yozzii last updated on 12/Apr/16
(dy/dx)=1−y^2 ⇒∫(dy/(1−y^2 ))=∫dx (1/2)∫(1/(1−y))+(1/(1+y))dy=x+C (1/2)(ln∣y+1∣−ln∣1−y∣)=x+C ((y+1)/(1−y))=e^(2x+2C) =Ae^(2x) y+1=Ae^(2x) −yAe^(2x) y(1+Ae^(2x) )=Ae^(2x) −1 y=((Ae^(2x) −1)/(Ae^(2x) +1))
$$\frac{{dy}}{{dx}}=\mathrm{1}−{y}^{\mathrm{2}} \Rightarrow\int\frac{{dy}}{\mathrm{1}−{y}^{\mathrm{2}} }=\int{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{1}−{y}}+\frac{\mathrm{1}}{\mathrm{1}+{y}}{dy}={x}+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\mid{y}+\mathrm{1}\mid−{ln}\mid\mathrm{1}−{y}\mid\right)={x}+{C} \\ $$$$\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}={e}^{\mathrm{2}{x}+\mathrm{2}{C}} ={Ae}^{\mathrm{2}{x}} \\ $$$${y}+\mathrm{1}={Ae}^{\mathrm{2}{x}} −{yAe}^{\mathrm{2}{x}} \\ $$$${y}\left(\mathrm{1}+{Ae}^{\mathrm{2}{x}} \right)={Ae}^{\mathrm{2}{x}} −\mathrm{1} \\ $$$${y}=\frac{{Ae}^{\mathrm{2}{x}} −\mathrm{1}}{{Ae}^{\mathrm{2}{x}} +\mathrm{1}} \\ $$$$ \\ $$
Answered by Yozzii last updated on 12/Apr/16
y′=1+((y/x))+((y/x))^2 (x≠0) Let y=ux⇒y^′ =u+u^′ x {u=f(x)} ∴ u+u^′ x=1+u+u^2 u^′ x=1+u^2 ⇒∫(1/(1+u^2 ))du=∫(1/x)dx tan^(−1) u=ln∣x∣+C ⇒u=tan(C+ln∣x∣) Since u=(y/x)⇒y=xtan(C+ln∣x∣) where C is constant.
$${y}'=\mathrm{1}+\left(\frac{{y}}{{x}}\right)+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \:\:\:\:\left({x}\neq\mathrm{0}\right) \\ $$$${Let}\:{y}={ux}\Rightarrow{y}^{'} ={u}+{u}^{'} {x}\:\:\:\:\left\{{u}={f}\left({x}\right)\right\} \\ $$$$\therefore\:{u}+{u}^{'} {x}=\mathrm{1}+{u}+{u}^{\mathrm{2}} \\ $$$${u}^{'} {x}=\mathrm{1}+{u}^{\mathrm{2}} \\ $$$$\Rightarrow\int\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}=\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$${tan}^{−\mathrm{1}} {u}={ln}\mid{x}\mid+{C} \\ $$$$\Rightarrow{u}={tan}\left({C}+{ln}\mid{x}\mid\right) \\ $$$${Since}\:{u}=\frac{{y}}{{x}}\Rightarrow{y}={xtan}\left({C}+{ln}\mid{x}\mid\right) \\ $$$${where}\:{C}\:{is}\:{constant}. \\ $$$$ \\ $$$$ \\ $$
Answered by Yozzii last updated on 12/Apr/16
yy^′ =−sinx⇒∫ydy=−∫sinxdx 0.5y^2 =−(−cosx)+C y^2 =2cosx+D y=±(√(D+2cosx)).
$${yy}^{'} =−{sinx}\Rightarrow\int{ydy}=−\int{sinxdx} \\ $$$$\mathrm{0}.\mathrm{5}{y}^{\mathrm{2}} =−\left(−{cosx}\right)+{C} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{cosx}+{D} \\ $$$${y}=\pm\sqrt{{D}+\mathrm{2}{cosx}}. \\ $$

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