MCQ
Solution of the differential equation, $y\,dx - x\,dy + x{y^2}dx = 0$ can be
- ✓$2x + {x^2}y = \lambda y$
- B$2y + {y^2}x = \lambda y$
- C$2y - {y^2}x = \lambda y$
- DNone of these
Integrating both side, we get $\frac{x}{y} = - \frac{{{x^2}}}{2} + c$
==> $2x + {x^2}y = 2cy$ ==> $2x + {x^2}y = \lambda y$ [$\lambda = 2c$]
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$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $-2$ : The equation in $\alpha $
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$