Let y=y(x) be the solution of the differentialequation {{dy}}/{{dx}}=1+{x} {e}^{{y}-{x}},-\\sqrt{2} < {x} < \\sqrt{2}, {y}(0)=0then, the minimum value of y(x), x \\in(-\\sqrt{2}, \\sqrt{2}) is equal to: [25 Jul 2021 Shift 1]

Correct Answer is : (1−√3)−loge(√3−1)

(1+\\sqrt{3})-\log _{e}(\\sqrt{3}-1)

(1-\\sqrt{3})-\log _{e}(\\sqrt{3}-1)

Differential Equations Part 1

Section: Mathematics

Question : 15 of 100

Marks: +1, -0

Let y=y(x) be the solution of the differentialequation

=1+xey−x,−√2<x<√2,y(0)=0then, the minimum value of y(x),x∈(−√2,√2) is equal to:

[25 Jul 2021 Shift 1]

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