Q. 39 - Calculus

Question

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{dy}{dx}=e^{x+y},y\left(0\right)=2$

Step-by-Step Solution

Answer

The solution of the initial-value problem descibed by$\frac{dy}{dx}=e^{x+y},y\left(0\right)=2$ is  $y\left(x\right)=-\ln |1+e^{-2}-e^x|$

Step 1. Given information

The given initial value problem $\frac{dy}{dx}=e^{x+y},y\left(0\right)=2..............\left(1\right)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value

Use the property of exponential function and rewrite the differential equation in (1) as
$\frac{dy}{dx}=e^x.e^y (\because e^x.e^y=e^{x+y})$
Note that the differentlal equation is now in the form $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$ In which $p\left(x\right)=e^x$, and $q\left(y\right)=e^y$. So, the differential equation can be solved by applying variable separable method. Thus, the solution of the differential equation involved in the initial- value problem is given by

$$\begin{gathered} \int \frac{1}{e^y}dy=\int e^{x}dx \\ \int e^{-y}dy=e^x+C \\ -e^{-y}=e^x+C \\ {e}^{-y}=-e^x-C \end{gathered}$$

Now, use the given initial condition $y\left(0\right)=2$, that is take $x=0,y=2$ in the above result and evaluate the constant C.

$$\begin{gathered} e^{-2}=-1-C \\ C=-e^{-2}-1 \end{gathered}$$

Substitute this value of the constant C in the solution of the differential equation and obtain the solution of the initial - value problem $\frac{dy}{dx}=e^{x+y},y\left(0\right)=2$ as 

$$\begin{gathered} e^{-y}=-e^x+1+e^{-2} \\ -y=\ln |1+e^{-2}-e^x| \\ y=-\ln |1+e^{-2}-e^x| \end{gathered}$$

Therefore,the solution of the initial-value problem descibed by equation(1) is

$y\left(x\right)=-\ln |1+e^{-2}-e^x|$