MCQ
The solution of $y\,dx - xdy + 3{x^2}{y^2}{e^{{x^3}}}dx = 0$ is
- ✓$\frac{x}{y} + {e^{{x^3}}} = c$
- B$\frac{x}{y} - {e^{{x^3}}} = 0$
- C$\frac{{ - x}}{y} + {e^{{x^3}}} = 0$
- DNone of these
$\frac{{ydx - xdy}}{{{y^2}}} + 3{x^2}{e^{{x^3}}}dx = 0$ ==> $d\left( {\frac{x}{y}} \right) + d{e^{{x^3}}} = 0$
On integrating, we get $\frac{x}{y} + {e^{{x^3}}} = c$
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$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$
$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix
$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$
$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix