Question

If $y=y\left(x\right)$ is the solution of the differential equation $\left(\frac{\left(5+e^x\right)}{\left(2+y\right)}\right)\frac{dy}{dx}+e^x=0$ satisfying $y\left(0\right)=1$ then the value of $y\left({\log }_{e}13\right)$ is.

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is D

$-1$

Explanation for the correct option:

Step 1: Finding the integration of the given differential equation.

Given,

$\begin{array}{rcl}\left(\frac{\left(5+e^x\right)}{\left(2+y\right)}\right)\frac{dy}{dx}+e^x& =& 0\\ & \Rightarrow & \left(\frac{\left(5+e^x\right)}{\left(2+y\right)}\right)\frac{dy}{dx}=-e^x\\ & \Rightarrow & \frac{dy}{\left(2+y\right)}=\left(\frac{-e^x}{\left(5+e^x\right)}\right)dx\end{array}$

Now integrate on both sides.

$\begin{array}{rcl}\int \frac{dy}{\left(2+y\right)}& =& \int \left(\frac{-e^x}{\left(5+e^x\right)}\right)dx\\ & & \end{array}$

$\begin{array}{rcl}& \Rightarrow & \ln \left(2+y\right)=-\ln \left(5+e^x\right)+\ln C\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{\int }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{}\mathbf{dx}\mathbf{}\mathbf{=}\mathbf{}\mathbf{ln}\mathbf{}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{C}\mathbf{]}...\left(1\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \ln \left(2+1\right)=-\ln \left(5+e^0\right)+\ln C\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{]}\\ & \Rightarrow & n\left(3\right)+\ln \left(6\right)=\ln C\\ & \Rightarrow & \ln \left(18\right)=\ln C\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{n}\mathbf{}\mathbf{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{l}\mathbf{n}\mathbf{}\mathbf{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{l}\mathbf{n}\mathbf{}\mathbf{a}\mathbf{b}\mathbf{]}\\ \mathbf{}\mathbf{}\therefore C& =& 18\end{array}$

Substituting the value of C in (1), we get;

$$\begin{gathered} \begin{array}{rcl}\ln \left(2+y\right)& =& -\ln \left(5+e^x\right)+\ln \left(18\right)\end{array} \\ \Rightarrow \ln (2+y)=\ln \left(\frac{18}{5+e^x}\right) \end{gathered}$$

Thus, the required equation is,

$\left(2+y\right)\left(5+e^x\right)=18$

Step 2 : Finding the value of $y\left({\log }_{e}13\right)$

Now put $x={\log }_{e}13$ to get the value of $y\left({\log }_{e}13\right)$

So,

$\begin{array}{rcl}\left(2+y\right)\left(5+e^{{\log }_{e}13}\right)& =& 18\\ \left(2+y\right)\left(5+13\right)& =& 18\left[\because a^{lo{g}_{a}b}=b\right]\\ \left(2+y\right)& =& 1\\ y& =& -1\end{array}$

Hence, the correct option is D.