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Question Number 96713 by Tony Lin last updated on 04/Jun/20

y^2  (d^2 y/dx^2 )=(dy/dx)

$${y}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{dy}}{{dx}} \\ $$

Answered by mr W last updated on 04/Jun/20

y^2  (d/dx)((dy/dx))=(dy/dx)  y^2  (d/dy)((dy/dx))(dy/dx)=(dy/dx)  (d/dy)((dy/dx))=(1/y^2 )  (dy/dx)=−((1/y)+C_1 )  ((ydy)/(1+C_1 y))=−dx  ∫((ydy)/(1+C_1 y))=−∫dx  y−∫(dy/(1+C_1 y))=−C_1 x+C_2   y−(1/C_1 )ln (1+C_1 y)=−C_1 x+C_2   ⇒y+C_1 x−(1/C_1 )ln (1+C_1 y)=C_2

$${y}^{\mathrm{2}} \:\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{dy}}{{dx}} \\ $$$${y}^{\mathrm{2}} \:\frac{{d}}{{dy}}\left(\frac{{dy}}{{dx}}\right)\frac{{dy}}{{dx}}=\frac{{dy}}{{dx}} \\ $$$$\frac{{d}}{{dy}}\left(\frac{{dy}}{{dx}}\right)=\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=−\left(\frac{\mathrm{1}}{{y}}+{C}_{\mathrm{1}} \right) \\ $$$$\frac{{ydy}}{\mathrm{1}+{C}_{\mathrm{1}} {y}}=−{dx} \\ $$$$\int\frac{{ydy}}{\mathrm{1}+{C}_{\mathrm{1}} {y}}=−\int{dx} \\ $$$${y}−\int\frac{{dy}}{\mathrm{1}+{C}_{\mathrm{1}} {y}}=−{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$${y}−\frac{\mathrm{1}}{{C}_{\mathrm{1}} }\mathrm{ln}\:\left(\mathrm{1}+{C}_{\mathrm{1}} {y}\right)=−{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{y}+{C}_{\mathrm{1}} {x}−\frac{\mathrm{1}}{{C}_{\mathrm{1}} }\mathrm{ln}\:\left(\mathrm{1}+{C}_{\mathrm{1}} {y}\right)={C}_{\mathrm{2}} \\ $$

Commented by Tony Lin last updated on 04/Jun/20

thanks sir

$${thanks}\:{sir} \\ $$

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