Question

Which of the following is the solution of the differential equation $\frac{dy}{dx}=e^{y+x}$ with initial condition $y\left(0\right)=-\ln 4$?

• A. $y=-x-\ln 4$
• B. $y=x-\ln 4$
• C. $y=-\ln \left(-e^x+5\right)$
• D. $y=-\ln \left(e^x+3\right)$
• E. $y=\ln \left(e^x+3\right)$

Answer 1

The correct option is C

$y=-\ln \left(-e^x+5\right)$

Explanation of correct option:

Given differential equation:

$\frac{dy}{dx}=e^{y+x}$

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=e^y·e^x\left[\because a^{m+n}=a^m·a^n\right] \\ \Rightarrow \frac{dy}{e^y}=e^{x}dx\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}{e}^y\right] \\ \Rightarrow e^{-y}dy=e^{x}dx\left[\because \frac{1}{a^m}=a^{-m}\right] \\ \therefore e^{-y}dy=e^{x}dx\dots \left(1\right) \end{gathered}$$

Integrating equation $\left(1\right)$

$\begin{array}{rcl}\int e^{-y}dy& =& \int e^{x}dx\\ & \Rightarrow & \frac{1}{-1}{e}^{-y}=e^x+C\left[\because \int e^{ax}dx=\frac{1}{a}{e}^x+c\right]\\ & \Rightarrow & -e^{-y}=e^x+C\\ & \Rightarrow & e^{-y}=-e^x-C\left[\mathrm{Multiply}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{by}\left(-1\right)\right]\\ & \Rightarrow & \ln \left(e^{-y}\right)=\ln \left(-e^x-C\right)\left[\mathrm{Taking}\mathrm{natural}\log \mathrm{on}\mathrm{both}\mathrm{sides}\right]\\ & \Rightarrow & -y=\ln \left(-e^x-C\right)\left[\because \ln \left(e^{ax}\right)=ax\right]\\ & \Rightarrow & y=-\ln \left(-e^x-C\right)\left[\mathrm{Multiply}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{by}\left(-1\right)\right]\end{array}$

$\therefore y=-\ln \left(-e^x-C\right)\dots \left(2\right)$

Where $C$ is constant of integration to be determined.

Given the initial condition is $y\left(0\right)=-\ln 4$.

Substitute $x=0$ and $y=-\ln 4$ in equation $\left(2\right)$.

$$\begin{gathered} -\ln 4=-\ln \left(-e^0-C\right) \\ \Rightarrow -\ln 4=-\ln \left(-1-C\right)\left[\because e^0=1\right] \\ \Rightarrow 4=-1-C\left[\mathrm{Taking}\mathrm{antilog}\mathrm{of}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow C=-1-4 \\ \therefore C=-5 \end{gathered}$$

Substitute $C=-5$ in equation $\left(2\right)$.

$$\begin{gathered} y=-\ln \left\{-e^x-\left(-5\right)\right\} \\ \therefore y=-\ln \left(-e^x+5\right) \end{gathered}$$

Hence, option (C) is correct.