The solution of d^2 y d x^2 = ^2 x + x e^x is - Vidyadip

MCQ

The solution of $\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}x + x{e^x}$ is

✓
$y = \log (\sec x) + (x - 2){e^x} + {c_1}x + {c_2}$
B
$y = \log (\sec x) + (x + 2){e^x} + {c_1}x + {c_2}$
C
$y = \log (\sec x) - (x + 2){e^x} + {c_1}x + {c_2}$
D
None of these

✓

Answer

Correct option: A.

$y = \log (\sec x) + (x - 2){e^x} + {c_1}x + {c_2}$

a
(a) $\frac{{{d^2}y}}{{d{x^2}}} = {\sec ^2}x + x{e^x}$

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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