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Question Number 87059 by jagoll last updated on 02/Apr/20

$${\left(\mathrm{y}{}^{\prime }\right)}^2-{\mathrm{xy}}^{\prime }+\mathrm{y}=0$$$$\mathrm{find}\mathrm{the}\mathrm{solution}$$

Answered by mr W last updated on 03/Apr/20

$$y^{\prime }=\frac{1}{2}(x\pm \sqrt{x^2-4y})$$$$letu=x^2-4y$$$$$$$$case1:u=constant=C$$$$\Rightarrow x^2-4y=-4C$$$$\Rightarrow y=\frac{x^2}{4}+C$$$$y^{\prime }=\frac{x}{2}$$$${\left(\frac{x}{2}\right)}^2-x\left(\frac{x}{2}\right)+\frac{x^2}{4}+C=0\Rightarrow C=0$$$$\Rightarrow y=\frac{x^2}{4}$$$$$$$$case2:u\ne constant$$$$\frac{du}{dx}=2x-4y^{\prime }$$$$2x-\frac{du}{dx}=2x\pm 2\sqrt{u}$$$$\frac{du}{dx}=\pm 2\sqrt{u}$$$$\int \frac{du}{2\sqrt{u}}=\pm \int dx$$$$\sqrt{u}=C\pm x$$$$u={(C\pm x)}^2=x^2+C(C\pm 2x)$$$$x^2-4y=x^2+C(C\pm 2x)$$$$\Rightarrow y=-\frac{C(C\pm 2x)}{4}$$$$$$$$check:$$$$y^{\prime }=-\frac{\pm C}{2}$$$${xy}^{\prime }=-\frac{\pm Cx}{2}$$$${xy}^{\prime }-y=-\frac{\pm 2Cx}{4}+\frac{C(C\pm 2x)}{4}=\frac{C^2}{4}={\left(y^{\prime }\right)}^2$$

Commented by jagoll last updated on 03/Apr/20

$$\mathrm{sir}\mathrm{y}=\frac{{\mathrm{x}}^2}{4}?$$

Commented by mr W last updated on 03/Apr/20

$$yes.y=\frac{x^2}{4}isalsoasolution,see$$$$above.$$

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