Question
$(x - {y^3})dx + 3x{y^2}dy = 0$ का हल है
${y^3} = t$ रखने पर, $dt = 3{y^2}dy$
$x\,dx - tdx + xdt = 0$ ==> $xdx + xdt - tdx = 0$
==> $\frac{{dx}}{x} + d\left( {\frac{t}{x}} \right) = 0$
समाकलन करने पर, $\log x + \frac{t}{x} = k$ ==> $\log x + \frac{{{y^3}}}{x} = k$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.