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Question Number 215958 by Tawa11 last updated on 22/Jan/25

$$\mathrm{Find}\mathrm{the}\mathrm{only}\mathrm{function}\mathrm{that}\mathrm{satisfy}$$$$\mathrm{the}\mathrm{expression}\mathrm{below}:$$$${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$$

Answered by Ghisom last updated on 23/Jan/25

$${\left(y^{\prime }\right)}^2=y^{″}$$$$u=y^{\prime }$$$$u^2=u^{\prime }$$$$u=-\frac{1}{x+C_1}\left[\mathrm{from}\mathrm{experience}\right]$$$$u^{\prime }=\frac{1}{{(x+C_1)}^2}=u^2$$$$y^{\prime }=-\frac{1}{x+C_1}\Rightarrow$$$$<y=-\int \frac{dx}{x+C_1}=C_2-\ln \mid x+C_1\mid$$

Commented by Tawa11 last updated on 03/Feb/25

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{apprecuate}.$$

Answered by mr W last updated on 25/Jan/25

$$say\frac{dy}{dx}=p$$$$p^2=\frac{dp}{dx}$$$$\int \frac{dp}{p^2}=\int dx$$$$-\frac{1}{p}=x+C_1$$$$\frac{dy}{dx}=p=-\frac{1}{x+C_1}$$$$\int dy=-\int \frac{dx}{x+C_1}$$$$\Rightarrow y=-\ln \mid x+C_1\mid +C_2$$

Commented by Tawa11 last updated on 03/Feb/25

$$\mathrm{Thanks}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

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