Question

Question: In Problems 1 - 30, solve the equation.

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

Short answer

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

Step-by-step solution

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.

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

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,

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

So, the solution is${\mathbf{e}}^x\mathbf{+}{\mathbf{ye}}^{-y}\mathbf{=}\mathit{C}$