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Question Number 126758 by bounhome last updated on 24/Dec/20

$$solve:$$$$1.y^{\prime 3}+y^2-y^{\prime 2}=0$$$$2.4y=x^2+y^{\prime 2}$$

Answered by liberty last updated on 24/Dec/20

$$\left(2\right){\left(\frac{dy}{dx}\right)}^2=4y-x^2$$$$\frac{dy}{dx}=\pm \sqrt{4y-x^2}$$$$lety=u^2{x}^2\to \frac{dy}{dx}=2{xu}^2+2x^{2}u\frac{du}{dx}$$$$\iff 2{xu}^2+2x^{2}u\frac{du}{dx}=\pm x\sqrt{4u^2-1}$$$$\iff 2xu\frac{du}{dx}=\pm \sqrt{4u^2-1}-2u^2$$$$\frac{2u}{\pm \sqrt{4u^2-1}-2u^2}du=\frac{dx}{x}$$$$\left(1\right)\int \frac{2u}{\sqrt{4u^2-1}-2u^2}du=\ln Cx$$$$let\tan w=\sqrt{4u^2-1}$$$$\int \frac{-2wdw}{w^2-2w+1}=\ln Cx$$$$-\int \frac{d(w^2-2w+1)}{w^2-2w+1}-\int \frac{2}{{(w-1)}^2}dw=\ln Cx$$$$-2\ln \mid w-1\mid +\frac{2}{w-1}=\ln Cx$$$$\frac{2}{{\tan }^{-1}\left(\sqrt{4u^2-1}\right)-1}=\ln (Cx\left({\tan }^{-1}{(\sqrt{4u^2-1}-1)}^2\right)$$$$\frac{2}{{\tan }^{-1}\left(\sqrt{\frac{4y-x^2}{x^2}}\right)-1}=\ln \left(Cx{({\tan }^{-1}\left(\sqrt{\frac{4y-x^2}{x^2}}\right)-1)}^2\right)$$

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