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Question Number 51709 by sinx last updated on 29/Dec/18

$$y^{″}=2y\left(y^{\prime }\right)$$$$$$$$solvetheequation$$

Answered by mr W last updated on 29/Dec/18

$$y^{″}=\frac{d\left(y^{\prime }\right)}{dx}=\frac{d\left(y^{\prime }\right)}{dy}\times \frac{dy}{dx}=y^{\prime }\frac{d\left(y^{\prime }\right)}{dy}$$$$y^{″}=2y\left(y^{\prime }\right)$$$$y^{\prime }\frac{d\left(y^{\prime }\right)}{dy}=2y\left(y^{\prime }\right)$$$$\frac{d\left(y^{\prime }\right)}{dy}=2y$$$$\Rightarrow y^{\prime }=y^2+c_1^2$$$$\Rightarrow \frac{dy}{dx}=y^2+c_1^2$$$$\Rightarrow \int \frac{dy}{y^2+c_1^2}=\int dx$$$$\Rightarrow \frac{1}{c_1}{\tan }^{-1}\frac{y}{c_1}=x+c_2$$$$\Rightarrow y=c_1\tan (c_{1}x+c_2)$$

Answered by Abdulhafeez Abu qatada last updated on 30/Dec/18

$$$$$$$$$$y^{″}=\frac{d}{dx}\left(y^2\right)$$$$\int y^{″}dx=y^2$$$$y^{\prime }+V=y^2$$$$y^{\prime }+V=y^2$$$$\frac{dy}{dx}=y^2+V$$$$\frac{dy}{y^2+V}=dx$$$$$$$$\frac{1}{C}{\tan }^{-1}\left(\frac{y}{C}\right)+C_2=x$$$$$$

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