Question Number 12228 by tawa last updated on 16/Apr/17
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2xy}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} } \\ $$
Answered by mrW1 last updated on 16/Apr/17
$${u}=\frac{{y}}{{x}} \\ $$$${y}={ux} \\ $$$$\frac{{dy}}{{dx}}={x}\frac{{du}}{{dx}}+{u} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\Rightarrow \\ $$$${x}\frac{{du}}{{dx}}+{u}=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${x}\frac{{du}}{{dx}}=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−{u}=\frac{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}=\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}=\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du}^{\mathrm{2}} =\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{w}}{{w}\left(\mathrm{1}−{w}\right)}{dw}=\int\frac{\mathrm{1}}{{x}}{dx}\:\:\:\:\:\:\:\left({w}={u}^{\mathrm{2}} \right) \\ $$$$\int\left[\frac{\mathrm{1}}{{w}}+\frac{\mathrm{2}}{\mathrm{1}−{w}}\right]{du}=\mathrm{2}\int\frac{\mathrm{1}}{{x}}{dx} \\ $$$$\mathrm{ln}\:{w}−\mathrm{2ln}\:\left(\mathrm{1}−{w}\right)=\mathrm{2ln}\:{x}+{C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{w}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{w}\right)^{\mathrm{2}} }={C}_{\mathrm{1}} \\ $$$$\mathrm{ln}\:\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }={C}_{\mathrm{1}} \\ $$$$\mathrm{2ln}\:\frac{{u}}{{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}={C}_{\mathrm{1}} \\ $$$$\frac{{u}}{{x}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}={e}^{\frac{{C}_{\mathrm{1}} }{\mathrm{2}}} =\frac{\mathrm{1}}{{C}} \\ $$$$\frac{{y}}{{x}^{\mathrm{2}} \left(\mathrm{1}−\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}=\frac{\mathrm{1}}{{C}} \\ $$$$\frac{{y}}{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{C}} \\ $$$${y}^{\mathrm{2}} +{Cy}={x}^{\mathrm{2}} \\ $$
Commented by tawa last updated on 16/Apr/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 16/Apr/17
$${error}\:{in}\:{line}\:\mathrm{6}. \\ $$
Commented by mrW1 last updated on 16/Apr/17
$${you}\:{are}\:{right}. \\ $$
Commented by tawa last updated on 16/Apr/17
$$\mathrm{please}\:\mathrm{sir},\:\mathrm{mrw}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{root}\:\mathrm{equation}\:\mathrm{formular}..\:\mathrm{i}\:\mathrm{need}\:\mathrm{the}\:\mathrm{image}\:\mathrm{again}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 16/Apr/17
$${let}\:{y}={tx}\:\:\:{or}\:\:\frac{{y}}{{x}}={t} \\ $$$$\frac{{dy}}{{dx}}={t}+{x}\frac{{dt}}{{dx}} \\ $$$$\Rightarrow\:{x}\frac{{dt}}{{dx}}\:=\:\frac{{dy}}{{dx}}−{t} \\ $$$${or}\:{x}\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{2}{x}\left({tx}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} {x}^{\mathrm{2}} }\:−{t} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:−\:{t} \\ $$$${x}\frac{{dt}}{{dx}}\:=\:\frac{{t}\left(\mathrm{2}−{t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:\frac{{dx}}{{x}}\:=−\:\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)}{dt} \\ $$$$\int\:\frac{{dx}}{{x}}\:=−\int\:\left(\:\frac{{A}}{{t}}+\frac{{B}}{{t}−\mathrm{1}}+\frac{{C}}{{t}+\mathrm{1}}\right){dt} \\ $$$${A}=−\mathrm{1}\:;\:{B}=\mathrm{1}\:;\:{C}=\mathrm{1}\:. \\ $$$${so},\:\mathrm{ln}\:{x}\:+\mathrm{ln}\:{k}=\:\mathrm{ln}\:{t}−\mathrm{ln}\:\left({t}−\mathrm{1}\right)−\mathrm{ln}\:\left({t}+\mathrm{1}\right) \\ $$$$\mathrm{ln}\:{kx}\:=\mathrm{ln}\:\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$${kx}\:=\frac{\left({y}/{x}\right)}{\left({y}/{x}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\boldsymbol{{kx}}\:=\:\frac{\boldsymbol{{xy}}}{\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }\:\: \\ $$$$\Rightarrow\:\boldsymbol{{x}}=\mathrm{0}\:{and}\:\:\frac{\boldsymbol{{y}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{y}}}\:=\:\frac{\mathrm{1}}{\boldsymbol{{k}}}\: \\ $$$${are}\:{solutions}\:{to}\:{the}\:{given} \\ $$$${differential}\:{eqn}. \\ $$
Commented by tawa last updated on 16/Apr/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$