MCQ
$(x - {y^3})dx + 3x{y^2}dy = 0$ નો ઉકેલ મેળવો.
- A$\log x + \frac{x}{y}$
- ✓$\log x + \frac{{{y^3}}}{x} = k$
- C$\log x - \frac{x}{{{y^3}}} = k$
- D$\log xy - {y^3} = k$
Put ${y^3} = t$ $ \Rightarrow $ $dt = 3{y^2}dy$
$x\,dx - tdx + xdt = 0$ ==> $xdx + xdt - tdx = 0$
==> $\frac{{dx}}{x} + d\left( {\frac{t}{x}} \right) = 0$
On integration, we get $\log x + \frac{t}{x} = k$ ==> $\log x + \frac{{{y^3}}}{x} = k$.
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