1. If \( \frac{dy}{dx} = 3x^2 - 1 \), and if \( y = -1 \) when \( x = 1 \), then \( y = \)






Answer is: option2

\( x^3 - x - 1 \)

Solution:

We are given the differential equation:

\( \frac{dy}{dx} = 3x^2 - 1 \)

and the initial condition:

\( y = -1 \) when \( x = 1 \)

\( \frac{dy}{dx} = 3x^2 - 1 \)

Integrating both sides with respect to \( x \):

\[ y = \int (3x^2 - 1)\, dx = x^3 - x + C \]

Given \( y = -1 \) when \( x = 1 \):

\[ -1 = (1)^3 - 1 + C \Rightarrow -1 = 1 - 1 + C \Rightarrow -1 = C \]

\[ y = x^3 - x - 1 \]

Final Answer: (B) \( x^3 - x - 1 \)

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