Chapter 7.4, Problem 7E

To determine

To solve initial value problem using variable separation.

Answer to Problem 7E

The solution of differential equation $\frac{dy}{dx}=\left(\cos x\right)e^{y+\sin x}$ is $y=-\ln \left(e+1-e^{\sin x}\right)$ and the domain over which the solution is valid is $\left(-\infty ,\infty \right)$ that is for all real values of $x$ .

Explanation of Solution

Given information:

The differential equation is $\frac{dy}{dx}=\left(\cos x\right)e^{y+\sin x}$ and the initial condition is $y=-1\text{ when }x=0$ .

Calculation:

The differential equation is $\frac{dy}{dx}=\left(\cos x\right)e^{y+\sin x}$

Using separation of variables method

  $\begin{array}{l}\frac{dy}{dx}=\left(\cos x\right)e^{y+\sin x}\\ \frac{dy}{dx}=\left(\cos x\right)e^y{e}^{\sin x}\\ \frac{1}{e^y}dy=e^{\sin x}\cos xdx\\ e^{-y}dy=e^{\sin x}\cos xdx\\ \text{integrating both sides }\\ {\displaystyle \int e^{-y}dy={\displaystyle \int e^{\sin x}\cos xdx}}\\ -e^{-y}=e^{\sin x}+A\\ e^{-y}=-e^{\sin x}-A\\ e^{-y}=C-e^{\sin x}\\ \text{Taking logarithm both side}\\ \text{ln}\left(e^{-y}\right)=\ln \left(C-e^{\sin x}\right)\\ -y=\ln \left(C-e^{\sin x}\right)\\ y=-\ln \left(C-e^{\sin x}\right)\\ \text{initial condition is }y=-1\text{ when }x=0\\ -1=-\ln \left(C-e^{\sin 0}\right)\\ 1=\ln \left(C-1\right)\\ e=C-1\\ C=e+1\\ y=-\ln \left(e+1-e^{\sin x}\right)\\ \text{domain : }\left(-\infty ,\infty \right)\end{array}$

Therefore, the solution of differential equation $\frac{dy}{dx}=\left(\cos x\right)e^{y+\sin x}$ is $y=-\ln \left(e+1-e^{\sin x}\right)$ and the domain over which the solution is valid is

  $\left(-\infty ,\infty \right)$ that is for all real values of $x$ .