MCQ
The general solution of the differential equation $(x + y)dx + xdy = 0$ is
- A${x^2} + {y^2} = c$
- B$2{x^2} - {y^2} = c$
- ✓${x^2} + 2xy = c$
- D${y^2} + 2xy = c$
==> $\frac{{dy}}{{dx}} = - \frac{{x + y}}{x}$
It is homogenous equation, hence put $y = vx$ and $\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}},$ we get $v + x\frac{{dv}}{{dx}} = - \frac{{x + vx}}{x} = - \frac{{1 + v}}{1}$
==> $x\frac{{dv}}{{dx}} = - 1 - 2v$==> $\int_{}^{} {\frac{{dv}}{{1 + 2v}}} = - \int_{}^{} {\frac{{dx}}{x}} $
==> $\frac{1}{2}\log (1 + 2v) = - \log x + \log c$ ==> $\log \left( {1 + 2\frac{y}{x}} \right) = 2\log \frac{c}{x}$
==> $\frac{{x + 2y}}{x} = {\left( {\frac{c}{x}} \right)^2}$ ==> ${x^2} + 2xy = c$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| X: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | a | 3a | 5a | 7a | 9a | 11a | 13a | 15a | 17a |
Match the conditions / expressions in Column $I$ with statements in Column $II$ and indicate your answers by darkening the appropriate bubbles in $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ If $-1 < x < 1$, then $f$ ( $x$ ) satisfies | $(p)$ $ 0 < $ f (x) $ < 1$ |
| $(B)$ If $1 < x < 2$, then $f(x)$ satisfies | $(q)$ $\mathrm{f}(\mathrm{x}) < 0$ |
| $(C)$ If $3 < x < 5$, then $f(x)$ satisfies | $(r)$ $ \mathrm{f}(\mathrm{x}) > 0$ |
| $(D)$ If $x > 5$, then $f(x)$ satisfies | $(s)$ $ f (\mathrm{x}) < 1$ |