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Expert-verified Found in: Page 79 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Question: In Problems 1 - 30, solve the equation.${\text{\hspace{0.17em}}}\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}\frac{{\mathbf{e}}^{x+y}}{y-1}$

${\mathbf{e}}^{x}+{\mathbf{ye}}^{-y}=C$

See the step by step solution

## Step 1: Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{e}}^{x+y}}{y-1}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Evaluate the equation (1).

$\begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{e}}^{x+y}}{y-1}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{e}}^{x}{\mathbf{e}}^{y}}{y-1}\\ \mathbf{=}\frac{{\mathbf{e}}^{x}}{\left(y-1\right){\mathbf{e}}^{-y}}\\ \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}\mathbf{=}{\mathbf{e}}^{x}\mathbf{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

Now integrate the equation (2) on both sides.

$\begin{array}{c}\int \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}\mathbf{=}\int {\mathbf{e}}^{x}\mathbf{dx}\\ \int \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}\mathbf{=}{\mathbf{e}}^{x}\mathbf{+}\mathbf{C}\end{array}$

## Step 2: Evaluation method

Find the value of $\int \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}$ separately.

Let us take $\mathbf{u}\mathbf{=}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{dv}\mathbf{=}{\mathbf{e}}^{-y}\mathbf{dy}$.

$\mathbf{du}\mathbf{=}\mathbf{1}\mathbf{dy}\mathbf{,}\mathbf{v}\mathbf{=}\mathbf{-}{\mathbf{e}}^{-y}$.

Use the integration by parts formula.

$\begin{array}{c}\int \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}\mathbf{=}\mathbf{-}{\mathbf{e}}^{-y}\left(y-1\right)\mathbf{+}\int {\mathbf{e}}^{-y}\mathbf{dy}\\ \mathbf{=}\mathbf{-}{\mathbf{ye}}^{-y}\mathbf{+}{\mathbf{e}}^{-y}\mathbf{-}{\mathbf{e}}^{-y}\\ \mathbf{=}\mathbf{-}{\mathbf{ye}}^{-y}\end{array}$

Then,

$\begin{array}{c}\int \left(y-1\right){\mathbf{e}}^{-y}\mathbf{dy}\mathbf{=}{\mathbf{e}}^{x}\mathbf{+}\mathbf{C}\\ \mathbf{-}{\mathbf{ye}}^{-y}\mathbf{=}{\mathbf{e}}^{x}\mathbf{+}\mathbf{C}\\ {\mathbf{e}}^{x}\mathbf{+}\text{ }{\mathbf{ye}}^{-y}=C\end{array}$

So, the solution is ${{\mathbf{e}}}^{x}{\mathbf{+}}{{\mathbf{ye}}}^{-y}{\mathbf{=}}{\mathbit{C}}$ ### Want to see more solutions like these? 