If y = y(x) is the solution of the differenti | vedclass

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(C) Given the differential equation: $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2)e^x = 0$.Rearranging the terms to separate variables:$\frac{dy}{1 + y^2}

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(C) Given the differential equation: $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2)e^x = 0$.Rearranging the terms to separate variables:$\frac{dy}{1 + y^2} = -\frac{2e^x}{1 + e^{2x}} dx$.Integrating both sides:$\int \frac{dy}{1 + y^2} = -\int \frac{2e^x}{1 + (e^x)^2} dx$.Let $u = e^x$,then $du = e^x dx$. The integral becomes:$	an^{-1}(y) = -2 	an^{-1}(e^x) + C$.Given $y(0) = 0$,substitute $x = 0$ and $y = 0$:$	an^{-1}(0) = -2 	an^{-1}(e^0) + C \implies 0 = -2(\frac{\pi}{4}) + C \implies C = \frac{\pi}{2}$.Thus,the solution is $	an^{-1}(y) = \frac{\pi}{2} - 2 	an^{-1}(e^x)$.To find $y'(0)$,substitute $x = 0$ into the original differential equation:$(1 + e^0) y'(0) + 2(1 + 0^2)e^0 = 0 \implies 2y'(0) + 2 = 0 \implies y'(0) = -1$.Now,find $y(\log_e \sqrt{3})$:$	an^{-1}(y) = \frac{\pi}{2} - 2 	an^{-1}(e^{\log_e \sqrt{3}}) = \frac{\pi}{2} - 2 	an^{-1}(\sqrt{3}) = \frac{\pi}{2} - 2(\frac{\pi}{3}) = \frac{\pi}{2} - \frac{2\pi}{3} = -\frac{\pi}{6}$.So,$y = 	an(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$.Then $(y(\log_e \sqrt{3}))^2 = (-\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.Finally,$6(y'(0) + (y(\log_e \sqrt{3}))^2) = 6(-1 + \frac{1}{3}) = 6(-\frac{2}{3}) = -4$.

If $y = y(x)$ is the solution of the differential equation $(1 + e^{2x}) \frac{dy}{dx} + 2(1 + y^2)e^x = 0$ and $y(0) = 0$,then $6(y'(0) + (y(\log_e \sqrt{3}))^2)$ is equal to

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