Let y(x) be a solution of the differential eq | vedclass

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(A) The given differential equation is $(1+e^x) y^{\prime}+y e^x=1$. Dividing by $(1+e^x)$,we get $\frac{d y}{d x}+\frac{e^x}{1+e^x} y = \frac{1}{1+e^

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$(A, C)$

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(A) The given differential equation is $(1+e^x) y^{\prime}+y e^x=1$. Dividing by $(1+e^x)$,we get $\frac{d y}{d x}+\frac{e^x}{1+e^x} y = \frac{1}{1+e^x}$.This is a linear differential equation of the form $\frac{d y}{d x}+P(x)y=Q(x)$.The integrating factor is $I.F. = e^{\int \frac{e^x}{1+e^x} dx} = e^{\ln(1+e^x)} = 1+e^x$.The solution is $y(1+e^x) = \int 1 dx = x+c$.Given $y(0)=2$,we have $2(1+e^0) = 0+c \Rightarrow c=4$.Thus,$y(x) = \frac{x+4}{1+e^x}$.For $(A)$,$y(-4) = \frac{-4+4}{1+e^{-4}} = 0$. So $(A)$ is true.For $(B)$,$y(-2) = \frac{-2+4}{1+e^{-2}} = \frac{2}{1+e^{-2}} 
eq 0$. So $(B)$ is false.For $(C)$ and $(D)$,we find the critical points by setting $y^{\prime}(x) = 0$.$y^{\prime}(x) = \frac{(1+e^x) - (x+4)e^x}{(1+e^x)^2} = \frac{1+e^x - xe^x - 4e^x}{(1+e^x)^2} = \frac{1-e^x(x+3)}{(1+e^x)^2}$.Let $g(x) = 1-e^x(x+3)$.$g(0) = 1-e^0(3) = 1-3 = -2$.$g(-1) = 1-e^{-1}(2) = 1-\frac{2}{e} \approx 1-0.736 = 0.264$.Since $g(x)$ is continuous and $g(-1) > 0$ and $g(0) Thus,$y(x)$ has a critical point in $(-1, 0)$. So $(C)$ is true.The correct options are $(A)$ and $(C)$.

Let $y(x)$ be a solution of the differential equation $(1+e^x) y^{\prime}+y e^x=1$. If $y(0)=2$,then which of the following statements is (are) true?
$(A)$ $y(-4)=0$
$(B)$ $y(-2)=0$
$(C)$ $y(x)$ has a critical point in the interval $(-1,0)$
$(D)$ $y(x)$ has no critical point in the interval $(-1,0)$

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Let $y(x)$ be a solution of the differential equation $(1+e^x) y^{\prime}+y e^x=1$. If $y(0)=2$,then which of the following statements is (are) true?$(A)$ $y(-4)=0$$(B)$ $y(-2)=0$$(C)$ $y(x)$ has a critical point in the interval $(-1,0)$$(D)$ $y(x)$ has no critical point in the interval $(-1,0)$