Let xdydx−y=x2(xex+ex−1)∀x∈R−{0} such that y(1)=e−1 , then y(2) is equal to___

xdy−ydxx2=(xex+ex−1)dx yx=xex−x+c x=1⇒y=e−1⇒c=0 y=x2(ex−1)

Let xdydx−y=x2(xex+ex−1)∀x∈R−{0} such that y(1)=e−1 , then y(2) is equal to___

xdy−ydxx2=(xex+ex−1)dx yx=xex−x+c x=1⇒y=e−1⇒c=0 y=x2(ex−1)

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Let $\frac{xdy}{dx} - y = x^{2} ({xe}^{x} + e^{x} - 1) \forall x \in R - {0}$ such that $y (1) = e - 1$ , then $y (2)$ is equal to___

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$4 (e - 1)$

$4 (e^{2} - 1)$

$2 (e^{2} - 1)$

$4 (e^{2} + 1)$

answer is B.

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Detailed Solution

$\frac{x d y - y d x}{x^{2}} = (x e^{x} + e^{x} - 1) d x \frac{y}{x} = x e^{x} - x + c x = 1 \Rightarrow y = e - 1 \Rightarrow c = 0 y = x^{2} (e^{x} - 1)$

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Let xdydx−y=x2(xex+ex−1)∀x∈R−{0} such that y(1)=e−1 , then y(2) is equal to___

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Let $\frac{xdy}{dx} - y = x^{2} ({xe}^{x} + e^{x} - 1) \forall x \in R - {0}$ such that $y (1) = e - 1$ , then $y (2)$ is equal to___