Question 76

Let $$y = y(x)$$ be the solution curve of the differential equation $$\sec y \frac{dy}{dx} + 2x\sin y = x^3\cos y$$, $$y(1) = 0$$. Then $$y(\sqrt{3})$$ is equal to :

$$\sec^2 y \frac{dy}{dx} + 2x \tan y = x^3$$.

Substitute $$v = \tan y \implies \frac{dv}{dx} + 2xv = x^3$$.

$$IF = e^{x^2}$$.

$$v e^{x^2} = \int x^3 e^{x^2} dx = \frac{1}{2}(x^2-1)e^{x^2} + C$$.

Using $$y(1)=0$$, $$C=0$$.

At $$x = \sqrt{3}$$: $$\tan y = \frac{1}{2}(3-1) = 1 \implies y = \frac{\pi}{4}$$.

Answer: D

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