x2y3+xydydx=1,y(1)=0,y=2 when x=

dxdy=xy+x2y3⇒−1x2dxdy+yx=−y3⇒ddy1x+1x⋅y=−y3I.F=exp⁡∫y⋅dy=ey2/2 Soln is ey2/2x=∫−y3ey2/2⋅dy⇒1x=2−y2+c.e−y2/2x=1,y=0⇒c=−1 ⇒x=−e

x2y3+xydydx=1,y(1)=0,y=2 when x=

dxdy=xy+x2y3⇒−1x2dxdy+yx=−y3⇒ddy1x+1x⋅y=−y3I.F=exp⁡∫y⋅dy=ey2/2 Soln is ey2/2x=∫−y3ey2/2⋅dy⇒1x=2−y2+c.e−y2/2x=1,y=0⇒c=−1 ⇒x=−e

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$(x^{2} y^{3} + x y) \frac{d y}{d x} = 1, y (1) = 0, y = \sqrt{2} when x =$

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-e

$\frac{- 1}{e}$

$\frac{1}{e}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{d x}{d y} = x y + x^{2} y^{3} \\ \Rightarrow \frac{- 1}{x^{2}} \frac{d x}{d y} + \frac{y}{x} = - y^{3} \end{array}$

$\begin{array}{l} \Rightarrow \frac{d}{d y} (\frac{1}{x}) + \frac{1}{x} \cdot y = - y^{3} \\ I . F = \exp \int y \cdot d y = e^{y^{2} / 2} \\ Soln is \frac{e^{y^{2} / 2}}{x} = \int - y^{3} e^{y^{2} / 2} \cdot d y \\ \Rightarrow \frac{1}{x} = 2 - y^{2} + c . e^{- y^{2} / 2} \\ x = 1, y = 0 \Rightarrow c = - 1 \Rightarrow x = - e \end{array}$