Chapter 2.RP, Problem 1RP

To determine

To solve:

The equation $\frac{dy}{dx}=\frac{e^{x+y}}{y-1}$.

Answer to Problem 1RP

Solution:

The solution of the equation $\frac{dy}{dx}=\frac{e^{x+y}}{y-1}$ is $e^x+y{e}^{-y}=c$.

Explanation of Solution

Formula used:

The formula of “Integration by parts” is given by,

${\displaystyle \int u\left(x\right)v\left(x\right)}dx=u\left(x\right){\displaystyle \int v\left(x\right)dx}-{\displaystyle \int \left[\frac{d}{dx}\left(u\left(x\right)\right){\displaystyle \int v\left(x\right)dx}\right]dx}$

Calculation:

Consider the given equation.

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

Separate the variable as follows,

$\begin{array}{rll}\frac{dy}{dx}& =& e^x\left(\frac{e^y}{y-1}\right)\\ \frac{y-1}{e^y}dy& =& e^{x}dx\\ \left(y-1\right)e^{-y}dy& =& e^{x}dx\\ y{e}^{-y}dy-e^{-y}dy& =& e^{x}dx\end{array}$

Integrate on both sides.

${\displaystyle \int y{e}^{-y}dy-{\displaystyle \int e^{-y}dy}}={\displaystyle \int e^{x}dx}\mathrm{...}(1)$

Solve ${\displaystyle \int y{e}^{-y}dy}$ by method of “Integration by parts”.

Let $u=y$, and $v=e^{-y}$.

Solve ${\displaystyle \int y{e}^{-y}dy}$ as follows,

$\begin{array}{rll}{\displaystyle \int y}{e}^{-y}dy& =& y{\displaystyle \int e^{-y}}dy-{\displaystyle \int \left[\frac{d}{dy}\left(y\right){\displaystyle \int e^{-y}}dy\right]}dy\\ & =& -y{e}^{-y}-{\displaystyle \int \left(-e^{-y}\right)}dy\\ & =& -y{e}^{-y}+{\displaystyle \int e^{-y}dy}\\ & =& -y{e}^{-y}-e^{-y}+C\end{array}$

Substitute the obtained results in equation $\left(1\right)$.

$\begin{array}{rll}-y{e}^{-y}-e^{-y}-{\displaystyle \int e^{-y}dy}& =& {\displaystyle \int e^{x}dx}\\ -y{e}^{-y}-e^{-y}-\left(-e^{-y}\right)+c& =& e^x\\ -y{e}^{-y}-e^{-y}+e^{-y}+c& =& e^x\\ -y{e}^{-y}+c& =& e^x\end{array}$

Further simplify,

$e^x+y{e}^{-y}=c$

Conclusion:

Thus, solution of the equation $\frac{dy}{dx}=\frac{e^{x+y}}{y-1}$ is $e^x+y{e}^{-y}=c$.