Given that dy/dx = yex such that x = 0, y = e. The value of y(y > 0) when x = 1 will be
e
1/e
ee
e2
The correct answer is option C) ee
Given that dy/dx = yex
such that x = 0, y = e.
The value of y(y > 0) when x = 1 will be
dy/y = exdx
⇒ ln y = ex + c
At x = 0, y = e.
So, c = 0. ln y = ex
Therefore, at x = 1, y = ee.
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For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: