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Question

If $$\frac { dy }{ dx } +xsin2y={ x }^{ 3 }{ cos }^{ 2 }y,y(0)=0,$$then y(1)equal to 


A
tan1(e)
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B
tan1(1e)
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C
tan1(e2)
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D
tan1(12e)
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Solution

The correct option is C $${ tan }^{ -1 }(e)$$
$$\dfrac{dy}{dx} + x \sin 2y = x^3 \cos^2 y$$
divide both sides by $$\cos^2 y$$
$$\sec^2 y \dfrac{dy}{dx} + \dfrac{2 \sin y \cos y}{\cos^2 y} x = x^3$$
$$\sec^2 y \dfrac{dy}{dx} + 2 \tan y . x = x^3$$
set $$\tan y = t \Rightarrow \sec^2 y \dfrac{dy}{d2} = \dfrac{d2}{dx}$$
$$\Rightarrow \dfrac{d2}{dx} + 2x . 2 = x^3$$
linear differential equation
$$I.F = e^{\int 2x dx} = e^{x^2}$$
$$\therefore$$ solution is $$t. e^{x^2} = \displaystyle \int e^{x^2} . x^3 \, dx + c$$
Let $$x^2 = n$$
$$x dx = \dfrac{1}{2} dn$$
$$\Rightarrow y . e^{x^2} = C + \displaystyle \int \dfrac{1}{2} xe^n dn$$
$$= C + \dfrac{1}{2} [ne^n - e^n] = C + \dfrac{e^n}{2} (n - 1)$$
$$\Rightarrow -\!\!\!\!z \, e^{x^2} = C + \dfrac{1}{2} e^{x^2} (x^2 - 1)$$
$$\Rightarrow \tan y = Ce^{-x^2} + \dfrac{1}{2} (x^2 -1)$$
$$\tan y = Ce^{-x^2} + \dfrac{1}{2} (x^2 - 1)$$

Mathematics

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