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Question Number 12228 by tawa last updated on 16/Apr/17

Solve the differential equation (dy/dx) = ((2xy)/(x^2 + y^2 ))

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2xy}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} } \\ $$

Answered by mrW1 last updated on 16/Apr/17

u=(y/x) y=ux (dy/dx)=x(du/dx)+u (dy/dx)=((2xy)/(x^2 +y^2 ))⇒ x(du/dx)+u=((2u)/(1+u^2 )) x(du/dx)=((2u)/(1+u^2 ))−u=((u(1−u^2 ))/(1+u^2 )) ((1+u^2 )/(u(1−u^2 )))du=(1/x)dx ∫((1+u^2 )/(u(1−u^2 )))du=∫(1/x)dx (1/2)∫((1+u^2 )/(u^2 (1−u^2 )))du^2 =∫(1/x)dx (1/2)∫((1+w)/(w(1−w)))dw=∫(1/x)dx (w=u^2 ) ∫[(1/w)+(2/(1−w))]du=2∫(1/x)dx ln w−2ln (1−w)=2ln x+C_1 ln (w/(x^2 (1−w)^2 ))=C_1 ln (u^2 /(x^2 (1−u^2 )^2 ))=C_1 2ln (u/(x(1−u^2 )))=C_1 (u/(x(1−u^2 )))=e^(C_1 /2) =(1/C) (y/(x^2 (1−(y^2 /x^2 ))))=(1/C) (y/(x^2 −y^2 ))=(1/C) y^2 +Cy=x^2

$${u}=\frac{{y}}{{x}} \\ $$$${y}={ux} \\ $$$$\frac{{dy}}{{dx}}={x}\frac{{du}}{{dx}}+{u} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\Rightarrow \\ $$$${x}\frac{{du}}{{dx}}+{u}=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${x}\frac{{du}}{{dx}}=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−{u}=\frac{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}=\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}=\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}^{\mathrm{2}} =\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{w}}{{w}\left(\mathrm{1}−{w}\right)}{dw}=\int\frac{\mathrm{1}}{{x}}{dx}\:\:\:\:\:\:\:\left({w}={u}^{\mathrm{2}} \right) \\ $$$$\int\left[\frac{\mathrm{1}}{{w}}+\frac{\mathrm{2}}{\mathrm{1}−{w}}\right]{du}=\mathrm{2}\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\mathrm{ln}\:{w}−\mathrm{2ln}\:\left(\mathrm{1}−{w}\right)=\mathrm{2ln}\:{x}+{C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{w}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{w}\right)^{\mathrm{2}} }={C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }={C}_{\mathrm{1}} \\ $$$$\mathrm{2ln}\:\frac{{u}}{{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}={C}_{\mathrm{1}} \\ $$$$\frac{{u}}{{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}={e}^{\frac{{C}_{\mathrm{1}} }{\mathrm{2}}} =\frac{\mathrm{1}}{{C}} \\ $$$$\frac{{y}}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{{C}} \\ $$$$\frac{{y}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{C}} \\ $$$${y}^{\mathrm{2}} +{Cy}={x}^{\mathrm{2}} \\ $$

Commented by tawa last updated on 16/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by ajfour last updated on 16/Apr/17

error in line 6.

$${error}\:{in}\:{line}\:\mathrm{6}. \\ $$

Commented by mrW1 last updated on 16/Apr/17

you are right.

$${you}\:{are}\:{right}. \\ $$

Commented by tawa last updated on 16/Apr/17

please sir, mrw the cubic root equation formular.. i need the image again sir

$$\mathrm{please}\:\mathrm{sir},\:\mathrm{mrw}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{root}\:\mathrm{equation}\:\mathrm{formular}..\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{image}\:\mathrm{again}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 16/Apr/17

let y=tx or (y/x)=t (dy/dx)=t+x(dt/dx) ⇒ x(dt/dx) = (dy/dx)−t or x(dt/dx) = ((2x(tx))/(x^2 +t^2 x^2 )) −t x(dt/dx)=((2t)/(t^2 +1)) − t x(dt/dx) = ((t(2−t^2 −1))/(t^2 +1)) ⇒ (dx/x) =− (((t^2 +1))/(t(t−1)(t+1)))dt ∫ (dx/x) =−∫ ( (A/t)+(B/(t−1))+(C/(t+1)))dt A=−1 ; B=1 ; C=1 . so, ln x +ln k= ln t−ln (t−1)−ln (t+1) ln kx =ln (t/(t^2 −1)) kx =(((y/x))/((y/x)^2 −1)) kx = ((xy)/(y^2 −x^2 )) ⇒ x=0 and ((y^2 −x^2 )/y) = (1/k) are solutions to the given differential eqn.

$${let}\:{y}={tx}\:\:\:{or}\:\:\frac{{y}}{{x}}={t} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$$\Rightarrow\:{x}\frac{{dt}}{{dx}}\:=\:\frac{{dy}}{{dx}}−{t} \\ $$$${or}\:{x}\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{2}{x}\left({tx}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} {x}^{\mathrm{2}} }\:−{t} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:−\:{t} \\ $$$${x}\frac{{dt}}{{dx}}\:=\:\frac{{t}\left(\mathrm{2}−{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:\frac{{dx}}{{x}}\:=−\:\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt} \\ $$$$\int\:\frac{{dx}}{{x}}\:=−\int\:\left(\:\frac{{A}}{{t}}+\frac{{B}}{{t}−\mathrm{1}}+\frac{{C}}{{t}+\mathrm{1}}\right){dt} \\ $$$${A}=−\mathrm{1}\:;\:{B}=\mathrm{1}\:;\:{C}=\mathrm{1}\:. \\ $$$${so},\:\mathrm{ln}\:{x}\:+\mathrm{ln}\:{k}=\:\mathrm{ln}\:{t}−\mathrm{ln}\:\left({t}−\mathrm{1}\right)−\mathrm{ln}\:\left({t}+\mathrm{1}\right) \\ $$$$\mathrm{ln}\:{kx}\:=\mathrm{ln}\:\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${kx}\:=\frac{\left({y}/{x}\right)}{\left({y}/{x}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\boldsymbol{{kx}}\:=\:\frac{\boldsymbol{{xy}}}{\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }\:\: \\ $$$$\Rightarrow\:\boldsymbol{{x}}=\mathrm{0}\:{and}\:\:\frac{\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{y}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{{k}}}\: \\ $$$${are}\:{solutions}\:{to}\:{the}\:{given} \\ $$$${differential}\:{eqn}. \\ $$

Commented by tawa last updated on 16/Apr/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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