((x^2 + 1)2dy + (y(2x^3 + x) – 2)dx = 0, y(0) = 0,) then y(2... | Mathematics | SolveeIt

$(x^2 + 1)2dy + (y(2x^3 + x) – 2)dx = 0, y(0) = 0,$ then y(2) is equal to

$\frac{2}5 tan^{-1}2$

$\frac{3}5 tan^{-1}2$

$\frac{2}5 tan^{-1}3$

$\frac{3}5 tan^{-1}3$

Answer

$\frac{2}5 tan^{-1}2$

Explanation

The Correct answer is option(A) : $\frac{2}5 tan^{-1}2$

Mathematics Question on Derivatives