Question
$\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}x + x{e^x}$ का हल है
समाकलन करने पर, $\frac{{dy}}{{dx}} = \tan x + x{e^x} - {e^x} + {c_1}$
पुन: $y = \log (\sec x) + x{e^x} - {e^x} - {e^x} + {c_1}x + {c_2}$
अत: अभीष्ट हल
$y = \log (\sec x) + (x - 2){e^x} + {c_1}x + {c_2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.