MCQ
The solution of the equation $\frac{{dy}}{{dx}} = \frac{1}{{x + y + 1}}$ is
- ✓$x = c{e^y} - y - 2$
- B$y = x + c{e^y} - 2$
- C$x + c{e^y} - y - 2 = 0$
- DNone of these
==>$\frac{{dx}}{{dy}} - x = y + 1$
It is linear equation, therefore $I.F. $$ = {e^{\int_{}^{} { - 1dy} }} = {e^{ - y}}$
Hence the solution of the equation is
$x.{e^{ - y}} = \int_{}^{} {(y + 1){e^{ - y}}} dy + c$ ==> $x = c{e^y} - y - 2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu} = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$
is a continuous function then the value of $k$ is