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Question Number 92114 by mhmd last updated on 04/May/20

Answered by mr W last updated on 05/May/20

$$u=y^{\prime }$$$$u\frac{du}{dy}+3y^{2}u=0$$$$\Rightarrow u=0\Rightarrow \frac{dy}{dx}=0\Rightarrow y=C$$$$\Rightarrow \frac{du}{dy}+3y^2=0\Rightarrow u=-(y^3-C^3)$$$$\Rightarrow \frac{dy}{dx}=-(y^3-C^3)$$$$\Rightarrow \frac{dy}{y^3-C^3}=-dx$$$$\Rightarrow \int \frac{dy}{y^3-C^3}=-x$$$$\Rightarrow \frac{1}{3C^2}\int (\frac{1}{y-C}-\frac{y+2C}{y^2+Cy+C^2})dy=-x$$$$\Rightarrow \frac{1}{3C^2}\int (\frac{1}{y-C}-\frac{1}{2}\times \frac{2y+C}{y^2+Cy+C^2}-\frac{3C}{2}\times \frac{1}{y^2+Cy+C^2})dy=-x$$$$...thisshouldbeeasy$$$$\Rightarrow x+\frac{1}{6C^2}\ln \mid \frac{{(y-C)}^2}{y^2+Cy+C^2}\mid -\frac{1}{\sqrt{3}{C}^2}{\tan }^{-1}\frac{2y+C}{\sqrt{3}C}+C_1=0$$

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