Question Id : 12875 |
Context : JEE Main 2022 (25 June Morning Shift)
Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$, with $y(0) = {1 \over 3}$. Then, the point $x = - {4 \over 3}$ for the curve $y = y(x)$ is :
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