MCQ
Solution of $ydx - xdy = {x^2}ydx$ is
- A$y{e^{{x^2}}} = c{x^2}$
- B$y{e^{ - {x^2}}} = c{x^2}$
- ✓${y^2}{e^{{x^2}}} = c{x^2}$
- D${y^2}{e^{ - {x^2}}} = c{x^2}$
After integration, we get $\log x - \frac{{{x^2}}}{2} = \log y + \log c$
==> $\log {x^2} - \log {y^2} + \log c = {x^2}$ ==> $\log \frac{{c{x^2}}}{{{y^2}}} = {x^2}$
==> $\frac{{c{x^2}}}{{{y^2}}} = {e^x}^2$ ==> $c{x^2} = {y^2}{e^{{x^2}}}$.
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Let $\mathrm{R}$ be a relation on $\mathrm{A} \times \mathrm{B}$ define by $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ if and only if $3 \mathrm{ad}-7 \mathrm{bc}$ is an even integer. Then the relation $\mathrm{R}$ is