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
    $x + c{e^y} - y - 2 = 0$
  • D
    None of these

Answer

Correct option: A.
$x = c{e^y} - y - 2$
a
(a) $\frac{{dy}}{{dx}} = \frac{1}{{x + y + 1}}$==> $\frac{{dx}}{{dy}} = x + y + 1$

==>$\frac{{dx}}{{dy}} - x = y + 1$

It is linear equation, therefore $I.F. $$ = {e^{\int_{}^{} { - 1dy} }} = {e^{ - y}}$

Hence the solution of the equation is

$x.{e^{ - y}} = \int_{}^{} {(y + 1){e^{ - y}}} dy + c$ ==> $x = c{e^y} - y - 2$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $9$ distinct balls be distributed among $4$ boxes, $B_{1}, B_{2}, B_{3}$ and $B_{4}$. If the probability that $B_{3}$ contains exactly $3$ balls is $k\left(\frac{3}{4}\right)^{9}$ then $\mathrm{k}$ lies in the set:
$10$ persons are seated around a round table. What is the probability that $4$ particular persons are always seated together?
If the system of linear equations  $x_1 + 2x_2 + 3x_3 = 6$ ; $x_1 + 3x_2 + 5x_3 = 9$ ; $2x_1 + 5x_2 + ax_3 = b$ is consistent and has infinite number of solutions, then
$i\,.\,(j \times k) + j\,.\,(k \times i) + k\,.\,(i \times j) = $
The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k}(\hat{i} \times \hat{j})$ is
The area bounded by the curve $\text{y}=\sec^2\text{x},\text{y}$ and $\text{x}=\frac{\pi}{3}$ is:
The figure formed by the four points $i + j - k,\,\,\,2i + 3j,$ $3i + 5j - 2k$ and $k - j$ is
Let $\lambda \in R , \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}$ If $((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}$, then $|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2$ is equal to
If $\left[ {\begin{array}{*{20}{c}}1&{\,\,1}&{\,\,1}\\1&{ - 2}&{ - 2}\\1&{\,\,3}&{\,\,1}\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$, then $\left[ \begin{array}{l}x\\y\\z\end{array} \right]$is equal to
Corner points of the bounded feasible region for an $ LP$ problem are $A(0, 5)\ B(0, 3)\ C(1, 0)\ D(6, 0).$ Let $z = -50x + 20y$ be the objective function. Minimum value of $z$ occurs at $.......$ center point.