If x y^{prime}+y-e^x=0 and y(a)=b,then lim _{ | vedclass

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(D) Given the differential equation: $x y^{\prime}+y-e^x=0$ This can be rewritten as: $x \frac{d y}{d x}+y=e^x$ Recognizing the product rule,we have:

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$e+a b-e^a$ (where $y^{\prime}=\frac{d y}{d x}$)

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(D) Given the differential equation: $x y^{\prime}+y-e^x=0$ This can be rewritten as: $x \frac{d y}{d x}+y=e^x$ Recognizing the product rule,we have: $\frac{d}{d x}(x y)=e^x$ Integrating both sides with respect to $x$: $\int d(x y)=\int e^x d x$ This yields: $x y=e^x+C$ Using the initial condition $y(a)=b$: $a b=e^a+C \Rightarrow C=a b-e^a$ Substituting $C$ back into the equation: $x y=e^x+a b-e^a$ Thus,$y(x)=\frac{e^x+a b-e^a}{x}$ Finally,evaluating the limit: $\lim _{x \rightarrow 1} y(x)=\frac{e^1+a b-e^a}{1}=e+a b-e^a$

If $x y^{\prime}+y-e^x=0$ and $y(a)=b$,then $\lim _{x \rightarrow 1} y(x)$ is

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