MCQ
${e^{dy/dx}} = (x + 1)$, $y(0) = 3$ નો ઉકેલ મેળવો.
- A$y = x\log x - x + 2$
- ✓$y = (x + 1)\log |x + 1| - x + 3$
- C$y = (x + 1)\log |x + 1| + x + 3$
- D$y = x\log x + x + 3$
$y = \int {\log (x + 1)dx} = x.\log (x + 1) - \int {\frac{x}{{x + 1}}dx} $
$ = x.\log (x + 1) - \int {\left( {1 - \frac{1}{{x + 1}}} \right)} \,dx$
$ = x.\log (x + 1) - x + \log (x + 1) + c$
$= (x + 1)\log (x + 1) - x + c$
$x = 0$ at $y = 3$
$3 = (1)\log (1) - 0 + c$ ==> $3 = 0 + c$ ==> $c = 3$
$y = (x + 1)\log |x + 1| - x + 3$.
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