If 1+xdydx-xy=1, y(0)=-1 then y(1)=

Given dydx−(x1+x)y=1x+1I.F=e∫−xx+1 dx=e∫(1x+1−1)=e log (x+1)−x=e log (x+1)e−x= e−x(x+1)Solution isy(I.F)= ∫1x+1(I.F) dx⇒ y e−x(x+1)= ∫1x+1e−x(x+1) dx=∫e−x dx=−e−x+c⇒ y(x+1)= −1+cexPut x=0⇒ y(0)(1)=−1+ce0⇒ −1=−1+c⇒ c=0y(x+1)=−1Put x=1⇒ y(1)(2)=−1⇒y(1)=−12

If 1+xdydx-xy=1, y(0)=-1 then y(1)=

Given dydx−(x1+x)y=1x+1I.F=e∫−xx+1 dx=e∫(1x+1−1)=e log (x+1)−x=e log (x+1)e−x= e−x(x+1)Solution isy(I.F)= ∫1x+1(I.F) dx⇒ y e−x(x+1)= ∫1x+1e−x(x+1) dx=∫e−x dx=−e−x+c⇒ y(x+1)= −1+cexPut x=0⇒ y(0)(1)=−1+ce0⇒ −1=−1+c⇒ c=0y(x+1)=−1Put x=1⇒ y(1)(2)=−1⇒y(1)=−12

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If $(1 + x) \frac{dy}{dx} - xy = 1, y (0) = - 1 then y (1) =$

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$\frac{1}{\sqrt{e}}$

1/2

$\sqrt{e}$

–1/2

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n \frac{d y}{d x} - (\frac{x}{1 + x}) y = \frac{1}{x + 1} \\ I . F = e^{\int \frac{- x}{x + 1} d x} = e^{\int (\frac{1}{x + 1} - 1)} \\ = e^{\log (x + 1) - x} \\ = e^{\log (x + 1)} e^{- x} \\ = e^{- x} (x + 1) \\ S o l u t i o n i s \\ y (I . F) = \int \frac{1}{x + 1} (I . F) d x \\ \Rightarrow y e^{- x} (x + 1) = \int \frac{1}{x + 1} e^{- x} (x + 1) d x \\ = \int e^{- x} d x \\ = - e^{- x} + c \end{array}$

$\begin{array}{l} \Rightarrow y (x + 1) = - 1 + c e^{x} \\ P u t x = 0 \\ \Rightarrow y (0) (1) = - 1 + c e^{0} \\ \Rightarrow - 1 = - 1 + c \\ \Rightarrow c = 0 \\ y (x + 1) = - 1 \\ P u t x = 1 \\ \Rightarrow y (1) (2) = - 1 \\ \Rightarrow y (1) = \frac{- 1}{2} \end{array}$