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Question Number 68113 by Joel122 last updated on 05/Sep/19

Solve  y.y′′ = 3(y′)^2

$$\mathrm{Solve} \\ $$$${y}.{y}''\:=\:\mathrm{3}\left({y}'\right)^{\mathrm{2}} \\ $$

Answered by Smail last updated on 05/Sep/19

⇔((y′′)/(y′))=3((y′)/y)  ln∣y′∣=3ln∣y∣+c  y′=ky^3   ((y′)/y^3 )=k⇔−(1/(2y^2 ))=kt+C  y^2 =((−2)/(kt+C))

$$\Leftrightarrow\frac{{y}''}{{y}'}=\mathrm{3}\frac{{y}'}{{y}} \\ $$$${ln}\mid{y}'\mid=\mathrm{3}{ln}\mid{y}\mid+{c} \\ $$$${y}'={ky}^{\mathrm{3}} \\ $$$$\frac{{y}'}{{y}^{\mathrm{3}} }={k}\Leftrightarrow−\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} }={kt}+{C} \\ $$$${y}^{\mathrm{2}} =\frac{−\mathrm{2}}{{kt}+{C}} \\ $$

Commented by Joel122 last updated on 05/Sep/19

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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