The solution of dy dx = e^x ( ^2 x + 2x) y(2 y + 1) is - Vidyadip

MCQ

The solution of $\frac{{dy}}{{dx}} = \frac{{{e^x}({{\sin }^2}x + \sin 2x)}}{{y(2\log y + 1)}}$ is

✓
${y^2}(\log y) - {e^x}{\sin ^2}x + c = 0$
B
${y^2}(\log y) - {e^x}{\cos ^2}x + c = 0$
C
${y^2}(\log y) + {e^x}{\cos ^2}x + c = 0$
D
None of these

✓

Answer

Correct option: A.

${y^2}(\log y) - {e^x}{\sin ^2}x + c = 0$

a
(a) $\frac{{dy}}{{dx}} = \frac{{{e^x}({{\sin }^2}x + \sin 2x)}}{{y(2\log y + 1)}}$

==>$\int_{}^{} {(2y\log y + y)dy = \int_{}^{} {{e^x}({{\sin }^2}x + \sin 2x} )dx} $

On integrating by parts, we get ${y^2}(\log y) = {e^x}{\sin ^2}x + c$.

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