MCQ
$x\frac{{dy}}{{dx}} = y(\log y - \log x + 1)$ નો ઉકેલ મેળવો.
- ✓$y = x{e^{cx}}$
- B$y + x{e^{cx}} = 0$
- C$y + {e^x} = 0$
- Dએકપણ નહી.
It is homogeneous equation
So now put $y=vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}$, then the equation $(i)$ reduces to$\frac{{dv}}{{v\log v}} = \frac{{dx}}{x}$
On integrating, we get $\log (\log v) = \log x + \log c$
==> $\log \left( {\frac{y}{x}} \right) = cx$ ==> $y = x{e^{cx}}$.
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