સમીકરણ dy dx = y^2 - y - 2 x^2 + 2x - 3 નો ઉકેલ મેળવો. - Vidyadip

MCQ

સમીકરણ $\frac{{dy}}{{dx}} = \frac{{{y^2} - y - 2}}{{{x^2} + 2x - 3}}$ નો ઉકેલ મેળવો.

A
$\frac{1}{3}\log \left| {\frac{{y - 2}}{{y + 1}}} \right| = \frac{1}{4}\log \left| {\frac{{x + 3}}{{x - 1}}} \right| + c$
B
$\frac{1}{3}\log \left| {\frac{{y + 1}}{{y - 2}}} \right| = \frac{1}{4}\log \left| {\frac{{x - 1}}{{x + 3}}} \right| + c$
✓
$4\log \left| {\frac{{y - 2}}{{y + 1}}} \right| = 3\log \left| {\frac{{x - 1}}{{x + 3}}} \right| + c$
D
એકપણ નહી.

✓

Answer

Correct option: C.

$4\log \left| {\frac{{y - 2}}{{y + 1}}} \right| = 3\log \left| {\frac{{x - 1}}{{x + 3}}} \right| + c$

c
(c) $\frac{{dy}}{{dx}} = \frac{{{y^2} - y - 2}}{{{x^2} + 2x - 3}}$ ==> $\frac{{dy}}{{(y - 2)(y + 1)}} = \frac{{dx}}{{(x + 3)(x - 1)}}$

==> $\int_{}^{} {\frac{{dy}}{{(y - 2)(y + 1)}}} = \int_{}^{} {\frac{{dx}}{{(x + 3)(x - 1)}}} $

==> $\frac{1}{3}\int_{}^{} {\left( {\frac{1}{{y - 2}} - \frac{1}{{y + 1}}} \right)} dy = \frac{1}{4}\int_{}^{} {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 3}}} \right)\,} dx$

==> $\frac{1}{3}\log \left| {\frac{{y - 2}}{{y + 1}}} \right| = \frac{1}{4}\left| {\frac{{x - 1}}{{x + 3}}} \right| + c$.

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