Question and Answers Forum

All Questions      Topic List

Differential Equation Questions

Previous in All Question      Next in All Question      

Previous in Differential Equation      Next in Differential Equation      

Question Number 103483 by bemath last updated on 15/Jul/20

y^2 −u(u+y). (dy/du) = 0

$${y}^{\mathrm{2}} −{u}\left({u}+{y}\right).\:\frac{{dy}}{{du}}\:=\:\mathrm{0} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Jul/20

y^2 =(dy/du).u(u+y)  y^2 =(dy/du)vy(vy+y)         {u=vy ,      1=(dv/du)y+(dy/du)v  1=(dy/du)v(v+1)  1=(1−(dv/du)y)(v+1)  1=v+1−(dv/du)y(v+1)  (v/(v+1))=(dv/du)((u/v))  ∫(du/u)=∫((v+1)/v^2 )dv  logu=logv−(1/v)+C  logy=−(y/u)+log(C_1 )  log((y/C_1 ))=−(y/u)  y=C_1 e^((−y)/u)

$${y}^{\mathrm{2}} =\frac{{dy}}{{du}}.{u}\left({u}+{y}\right) \\ $$$${y}^{\mathrm{2}} =\frac{{dy}}{{du}}{vy}\left({vy}+{y}\right)\:\:\:\:\:\:\:\:\:\left\{{u}={vy}\:,\:\:\:\:\:\:\mathrm{1}=\frac{{dv}}{{du}}{y}+\frac{{dy}}{{du}}{v}\right. \\ $$$$\mathrm{1}=\frac{{dy}}{{du}}{v}\left({v}+\mathrm{1}\right) \\ $$$$\mathrm{1}=\left(\mathrm{1}−\frac{{dv}}{{du}}{y}\right)\left({v}+\mathrm{1}\right) \\ $$$$\mathrm{1}={v}+\mathrm{1}−\frac{{dv}}{{du}}{y}\left({v}+\mathrm{1}\right) \\ $$$$\frac{{v}}{{v}+\mathrm{1}}=\frac{{dv}}{{du}}\left(\frac{{u}}{{v}}\right) \\ $$$$\int\frac{{du}}{{u}}=\int\frac{{v}+\mathrm{1}}{{v}^{\mathrm{2}} }{dv} \\ $$$${logu}={logv}−\frac{\mathrm{1}}{{v}}+{C} \\ $$$${logy}=−\frac{{y}}{{u}}+{log}\left({C}_{\mathrm{1}} \right) \\ $$$${log}\left(\frac{{y}}{{C}_{\mathrm{1}} }\right)=−\frac{{y}}{{u}} \\ $$$${y}={C}_{\mathrm{1}} {e}^{\frac{−{y}}{{u}}} \\ $$$$ \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 15/Jul/20

I have  started my lesson on Differential equations . Kindly  check my answer

$${I}\:{have}\:\:{started}\:{my}\:{lesson}\:{on}\:{Differential}\:{equations}\:.\:{Kindly} \\ $$$${check}\:{my}\:{answer} \\ $$

Answered by bramlex last updated on 15/Jul/20

(dy/du) = (y^2 /(u^2 +uy)) . set y = uw   (dy/du) = w + u(dw/du)  ⇔ w + u (dw/du) = ((u^2 w^2 )/(u^2 +u^2 w))  w + u (dw/du) = (w^2 /(1+w))  u (dw/du) = (w^2 /(1+w))−w = ((−w)/(1+w))  ((1+w )/w) dw = −(du/u)   ∫ ((1/w)+1)dw = −ln∣u∣+C  ln∣w∣+w = C−ln∣u∣  ln∣uw∣ +w = C  ln∣u.(y/u)∣+(y/u)=C  u ln∣y∣ +y = Cu  y = u(C−ln∣y∣) ■

$$\frac{\mathrm{dy}}{\mathrm{du}}\:=\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{u}^{\mathrm{2}} +\mathrm{uy}}\:.\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{uw}\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{du}}\:=\:\mathrm{w}\:+\:\mathrm{u}\frac{\mathrm{dw}}{\mathrm{du}} \\ $$$$\Leftrightarrow\:\mathrm{w}\:+\:\mathrm{u}\:\frac{\mathrm{dw}}{\mathrm{du}}\:=\:\frac{\mathrm{u}^{\mathrm{2}} \mathrm{w}^{\mathrm{2}} }{\mathrm{u}^{\mathrm{2}} +\mathrm{u}^{\mathrm{2}} \mathrm{w}} \\ $$$$\mathrm{w}\:+\:\mathrm{u}\:\frac{\mathrm{dw}}{\mathrm{du}}\:=\:\frac{\mathrm{w}^{\mathrm{2}} }{\mathrm{1}+\mathrm{w}} \\ $$$$\mathrm{u}\:\frac{\mathrm{dw}}{\mathrm{du}}\:=\:\frac{\mathrm{w}^{\mathrm{2}} }{\mathrm{1}+\mathrm{w}}−\mathrm{w}\:=\:\frac{−\mathrm{w}}{\mathrm{1}+\mathrm{w}} \\ $$$$\frac{\mathrm{1}+\mathrm{w}\:}{\mathrm{w}}\:\mathrm{dw}\:=\:−\frac{\mathrm{du}}{\mathrm{u}}\: \\ $$$$\int\:\left(\frac{\mathrm{1}}{\mathrm{w}}+\mathrm{1}\right)\mathrm{dw}\:=\:−\mathrm{ln}\mid\mathrm{u}\mid+\mathrm{C} \\ $$$$\mathrm{ln}\mid\mathrm{w}\mid+\mathrm{w}\:=\:\mathrm{C}−\mathrm{ln}\mid\mathrm{u}\mid \\ $$$$\mathrm{ln}\mid\mathrm{uw}\mid\:+\mathrm{w}\:=\:\mathrm{C} \\ $$$$\mathrm{ln}\mid\mathrm{u}.\frac{\mathrm{y}}{\mathrm{u}}\mid+\frac{\mathrm{y}}{\mathrm{u}}=\mathrm{C} \\ $$$$\mathrm{u}\:\mathrm{ln}\mid\mathrm{y}\mid\:+\mathrm{y}\:=\:\mathrm{Cu} \\ $$$$\mathrm{y}\:=\:\mathrm{u}\left(\mathrm{C}−\mathrm{ln}\mid\mathrm{y}\mid\right)\:\blacksquare\: \\ $$

Answered by OlafThorendsen last updated on 15/Jul/20

y = uY  u^2 Y^2 −u(u+uY)(u(dY/du)+Y) = 0  u^2 Y^2 −u^3 (1+Y)(dY/du)−u^2 (1+Y)Y = 0  −u^3 (1+Y)(dY/du)−u^2 Y = 0  ((1+Y)/Y)dY = −(du/u)  ln∣Y∣+Y = −ln∣u∣+C  ln∣u∣ = −ln∣Y∣−Y+C  ∣u∣ = ((Ke^(−Y) )/(∣Y∣)) = ((Ke^(−(y/u)) )/(∣(y/u)∣))  u^2  = ((Ke^(−(y/u)) )/y)

$${y}\:=\:{u}\mathrm{Y} \\ $$$${u}^{\mathrm{2}} \mathrm{Y}^{\mathrm{2}} −{u}\left({u}+{u}\mathrm{Y}\right)\left({u}\frac{{d}\mathrm{Y}}{{du}}+\mathrm{Y}\right)\:=\:\mathrm{0} \\ $$$${u}^{\mathrm{2}} \mathrm{Y}^{\mathrm{2}} −{u}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{Y}\right)\frac{{d}\mathrm{Y}}{{du}}−{u}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{Y}\right)\mathrm{Y}\:=\:\mathrm{0} \\ $$$$−{u}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{Y}\right)\frac{{d}\mathrm{Y}}{{du}}−{u}^{\mathrm{2}} \mathrm{Y}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{1}+\mathrm{Y}}{\mathrm{Y}}{d}\mathrm{Y}\:=\:−\frac{{du}}{{u}} \\ $$$$\mathrm{ln}\mid\mathrm{Y}\mid+\mathrm{Y}\:=\:−\mathrm{ln}\mid{u}\mid+\mathrm{C} \\ $$$$\mathrm{ln}\mid{u}\mid\:=\:−\mathrm{ln}\mid\mathrm{Y}\mid−\mathrm{Y}+\mathrm{C} \\ $$$$\mid{u}\mid\:=\:\frac{\mathrm{K}{e}^{−\mathrm{Y}} }{\mid\mathrm{Y}\mid}\:=\:\frac{\mathrm{K}{e}^{−\frac{{y}}{{u}}} }{\mid\frac{{y}}{{u}}\mid} \\ $$$${u}^{\mathrm{2}} \:=\:\frac{\mathrm{K}{e}^{−\frac{{y}}{{u}}} }{{y}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com