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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$\text{Let}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if,
$\theta=\frac{\pi}{4}$
$\theta=\frac{\pi}{3}$
$\theta=\frac{\pi}{2}$
$\theta=\frac{2\pi}{3}$