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Question Number 64973 by Tawa1 last updated on 23/Jul/19

Answered by mr W last updated on 23/Jul/19

(1)  (dy/dx)=u  (d^2 y/dx^2 )=(du/dx)=(du/dy)×(dy/dx)=u(du/dy)  u(du/dy)=(4/y^3 )  udu=((4dy)/y^3 )  ∫udu=∫((4dy)/y^3 )  (u^2 /2)=−(2/y^2 )+C_1   u^2 =((c_1 y^2 −4)/y^2 )  u=(dy/dx)=((√(c_1 y^2 −4))/y)  ((ydy)/(√(c_1 y^2 −4)))=dx  ∫((ydy)/(√(c_1 y^2 −4)))=∫dx  ∫((d(c_1 y^2 ))/(√(c_1 y^2 −4)))=2c_1 x+C_2   ⇒2(√(c_1 y^2 −4))=2c_1 x+C_2   ⇒(√(c_1 y^2 −4))=c_1 x+c_2

$$\left(\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}={u} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$${u}\frac{{du}}{{dy}}=\frac{\mathrm{4}}{{y}^{\mathrm{3}} } \\ $$$${udu}=\frac{\mathrm{4}{dy}}{{y}^{\mathrm{3}} } \\ $$$$\int{udu}=\int\frac{\mathrm{4}{dy}}{{y}^{\mathrm{3}} } \\ $$$$\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\mathrm{2}}{{y}^{\mathrm{2}} }+{C}_{\mathrm{1}} \\ $$$${u}^{\mathrm{2}} =\frac{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}{{y}^{\mathrm{2}} } \\ $$$${u}=\frac{{dy}}{{dx}}=\frac{\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}}{{y}} \\ $$$$\frac{{ydy}}{\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}}={dx} \\ $$$$\int\frac{{ydy}}{\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}}=\int{dx} \\ $$$$\int\frac{{d}\left({c}_{\mathrm{1}} {y}^{\mathrm{2}} \right)}{\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}}=\mathrm{2}{c}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}=\mathrm{2}{c}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{{c}_{\mathrm{1}} {y}^{\mathrm{2}} −\mathrm{4}}={c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \\ $$

Commented by Tawa1 last updated on 23/Jul/19

God bless you sir.

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

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