Consider a differential equation, it is given that corresponding curve passes through (0, 1) then value of y for x = 1 is
$\frac{d y}{d x} - e^{- x^{2}} + 2 x y = 0$
0.736

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Solution

The correct option is A 0.736
Given differential equation can be arranged as,
$\frac{d y}{d x} + 2 x y = e^{- x^{2}}$ which is a linear differential equation
Now integrating factor (IF),
$I . F = e^{\int 2 x d x} = e^{x^{2}}$
So solution of differential equation is
$y . e^{x^{2}} = \int e^{- x^{2}} e^{x^{2}} d x$
$\Rightarrow y e^{x^{2}} = x + c$
Now, given that y(0) = 1
$\Rightarrow 1. e^{0} = c$
$\Rightarrow c = 1$
$\Rightarrow y e^{x^{2}} = x + 1$
Now y(1) is given by,
$y . e^{1} = 1 + 1$
$\Rightarrow y (1) = \frac{2}{e} = 0.7358 ≃ 0.736$

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Similar questions

(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{(1 + x^{2})}

\frac{d y}{d x} = \frac{- 2 x y}{(x^{2} + 1)}

\frac{d y}{d x} = \frac{- 2 x y}{(x^{2} + 1)}

(x = 1, y = 0)

\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 y} + \frac{y}{x}

y = y (x)

\frac{d y}{d x} - \frac{2 x y}{1 + x^{2}} = 0

y (0) = 1,

y (1)

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