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MathematicsSingle correct+4 / -1Question 9 of 90

JEE Main 2022 · 29 Jun · Shift 2, Mathematics question 9

If y=y(x)y=y(x) is the solution of the differential equation (1+e2x)dydx+2(1+y2)ex=0\left(1+e^{2x}\right)\dfrac{dy}{dx}+2\left(1+y^2\right)e^x=0 and y(0)=0y(0)=0, then 6(y′(0)+(y(log⁡e3))2)6\left(y'(0)+\left(y\left(\log_e\sqrt3\right)\right)^2\right) is equal to

Choose one option
Answer and worked solution

Official answer: C

Separate. dy1+y2=−2ex1+e2xdx\dfrac{dy}{1+y^2}=-\dfrac{2e^x}{1+e^{2x}}dx: tan⁡−1y=π2−2tan⁡−1ex\tan^{-1}y=\frac\pi2-2\tan^{-1}e^x.

Values. y′(0)=−22⋅1=−1y'(0)=-\frac{2}{2}\cdot1=-1. At ex=3e^x=\sqrt3: tan⁡−1y=−π6\tan^{-1}y=-\frac\pi6, y2=13y^2=\frac13.

Answer. 6(−1+13)=−46\left(-1+\frac13\right)=-4.

(C)

Topics:Variable separable
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