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Question-100048




Question Number 100048 by bramlex last updated on 24/Jun/20
Answered by mr W last updated on 24/Jun/20
u=(dy/dx)  (d^2 y/dx^2 )=(du/dx)=(du/dy)×(dy/dx)=u(du/dy)  2yu(du/dy)−u^2 =1  ((2u)/(u^2 +1))du=(dy/y)  ∫((2u)/(u^2 +1))du=∫(dy/y)  ∫(1/(u^2 +1))d(u^2 +1)=∫(dy/y)  ln (u^2 +1)=ln y+C  ⇒u^2 +1=Cy  ⇒u=(dy/dx)=±(√(Cy−1))  ⇒(dy/( (√(Cy−1))))=±dx  ⇒∫(dy/( (√(Cy−1))))=±∫dx  (2/C)(√(Cy−1))=±x+C_1   ⇒2(√(Cy−1))=±C(x+C_1 )
$${u}=\frac{{dy}}{{dx}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$$\mathrm{2}{yu}\frac{{du}}{{dy}}−{u}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\frac{{dy}}{{y}} \\ $$$$\int\frac{\mathrm{2}{u}}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\int\frac{{dy}}{{y}} \\ $$$$\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}{d}\left({u}^{\mathrm{2}} +\mathrm{1}\right)=\int\frac{{dy}}{{y}} \\ $$$$\mathrm{ln}\:\left({u}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{ln}\:{y}+{C} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{1}={Cy} \\ $$$$\Rightarrow{u}=\frac{{dy}}{{dx}}=\pm\sqrt{{Cy}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{dy}}{\:\sqrt{{Cy}−\mathrm{1}}}=\pm{dx} \\ $$$$\Rightarrow\int\frac{{dy}}{\:\sqrt{{Cy}−\mathrm{1}}}=\pm\int{dx} \\ $$$$\frac{\mathrm{2}}{{C}}\sqrt{{Cy}−\mathrm{1}}=\pm{x}+{C}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\sqrt{{Cy}−\mathrm{1}}=\pm{C}\left({x}+{C}_{\mathrm{1}} \right) \\ $$

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