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$If \frac{d y}{d x} + x \sin 2 y = x^{3} \cos^{2} y, y (0) = 0 then \tan y (1) =$

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Correct option is B

sec2⁡ydydx+2xtan⁡y=x3 I. F=e∫2xdx=ex2 Solution is tan⁡yex2=∫x3⋅ex2dx=12x2−1ex2+c⇒tan⁡y(x)=12x2−1+c⋅e−x2x=0,y=0⇒c=12⇒tan⁡y(x)=12x2−1+e−x2tan⁡y(y(1))=12e

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If dydx+xsin⁡2y=x3cos2⁡y,y(0)=0 then tan⁡y(1)=

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$If \frac{d y}{d x} + x \sin 2 y = x^{3} \cos^{2} y, y (0) = 0 then \tan y (1) =$