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Question Number 107401 by mathmax by abdo last updated on 10/Aug/20

$$\mathrm{sove}{\mathrm{y}}^{{}^{″}}={\mathrm{y}}^2$$

Answered by mr W last updated on 10/Aug/20

$$y^{″}=y^{\prime }\frac{d\left(y^{\prime }\right)}{dy}=u\frac{du}{dy}$$$$u\frac{du}{dy}=y^2$$$$\Rightarrow \frac{u^2}{2}=\frac{y^3}{3}+C_1$$$$\Rightarrow u^2=\frac{2}{3}(y^3+C_1)$$$$\Rightarrow u=\frac{dy}{dx}=\pm \sqrt{\frac{2}{3}}\times \sqrt{y^3+C_1}$$$$\Rightarrow \pm \sqrt{\frac{3}{2}}\frac{dy}{\sqrt{y^3+C_1}}=dx$$$$\Rightarrow \pm \sqrt{\frac{3}{2}}\int \frac{dy}{\sqrt{y^3+C_1}}=\int dx$$$$\Rightarrow x=\pm \sqrt{\frac{3}{2}}\int \frac{dy}{\sqrt{y^3+C_1}}+C_2$$

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