MCQ
$\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}x + x{e^x}$ નો ઉકેલ મેળવો.
- ✓$y = \log (\sec x) + (x - 2){e^x} + {c_1}x + {c_2}$
- B$y = \log (\sec x) + (x + 2){e^x} + {c_1}x + {c_2}$
- C$y = \log (\sec x) - (x + 2){e^x} + {c_1}x + {c_2}$
- Dએકપણ નહી.
On integrating, $\frac{{dy}}{{dx}} = \tan x + x{e^x} - {e^x} + {c_1}$
Again, $y = \log (\sec x) + x{e^x} - {e^x} - {e^x} + {c_1}x + {c_2}$
Thus required solution is
$y = \log (\sec x) + (x - 2){e^x} + {c_1}x + {c_2}$.
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