The solution of $$2ycos$$⁡$$y2dydx$$−$$2x+1sin$$⁡$$y2=(x+1)3$$ is

Correct option is 2siny2=x+12x+122+c

Questions

$The solution of 2 y \cos y^{2} \frac{d y}{d x} - \frac{2}{x + 1} \sin y^{2} = (x + 1)^{3} is$

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siny2=x+12x+12+c

siny2=x+12x+122+c

siny2=x+12x+133+c

siny2=x+12x+144+c

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detailed solution

Correct option is B

Let sin⁡y2=t⇒2ycos⁡y2dydx=dtdx⇒ dtdx−2t(x+1)=(x+1)3 Solve the linear equation I. F=1(x+1)2G.S is t I.F.=∫Q(x) IFdxt(x+1)2=∫(x+1)dxsiny2(x+1)2=(x+1)22+c

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The solution of 2ycos⁡y2dydx−2x+1sin⁡y2=(x+1)3 is

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$The solution of 2 y \cos y^{2} \frac{d y}{d x} - \frac{2}{x + 1} \sin y^{2} = (x + 1)^{3} is$