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Mathematics · 2021

JEE Main · 27 July 2021, Shift 2 · Q68

Let y=y(x) be the solution of the differential equation (x-x^3) d y=(y+y x^2-3 x^4) d x, x>2. If y(3)=3, then y(4) is equal to:

Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle \left(x-x^3\right) \mathrm{d} y=\left(y+y x^2-3 x^4\right) \mathrm{d} x, x>2$. If $\displaystyle y(3)=3$, then $\displaystyle y(4)$ is equal to:
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