JEE MAIN - Mathematics Bengali (2022 - 24th June Evening Shift - No. 14)
যদি $$y = {\tan ^{ - 1}}\left( {\sec \,{x^3} - \tan \,{x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, হয় তবে
$$xy'' + 2y' = 0$$
$${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$
$${x^2}y'' - 6y + 3\pi = 0$$
$$xy'' - 4y' = 0$$
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