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Found in: Page 46

### 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

# In problem 7-16, solve the equation. $\frac{\mathrm{dx}}{\mathrm{dt}}{\mathbf{=}}\frac{t}{\mathbf{x}\text{ }{\mathbf{e}}^{t+2x}}$

The solution of the given differential equation is ${{\mathbf{e}}}^{2x}\left(2x-1\right){\mathbf{+}}{\mathbf{4}}{{\mathbf{e}}}^{-t}\left(t+1\right){\mathbf{=}}{\mathbf{C}}$.

See the step by step solution

## Step 1: Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{f}}\left(x,y\right)$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions ${\mathbf{g}}\left(x\right)$ that is a function of x alone and ${\mathbf{h}}\left(y\right)$ that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}{\mathbf{=}}{\mathbf{g}}\left(x\right){\text{ }}{\mathbf{h}}\left(y\right)$. By separating the variables, the equation follows $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}{\mathbf{=}}{\mathbf{g}}\left(x\right){\mathbf{dx}}$. Then, on direct integration of both sides, the solution of the differential equation is determined.

## Step 2: Solution of the Equation

The given equation is

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{t}}{\mathrm{x}\text{ }{\mathrm{e}}^{\mathrm{t}+2\mathrm{x}}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

$\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{t}}{\mathrm{x}\text{ }{\mathrm{e}}^{\mathrm{t}}\text{ }{\mathrm{e}}^{2\mathrm{x}}}\\ \mathrm{x}\text{ }{\mathrm{e}}^{2\mathrm{x}}\text{ }\mathrm{dx}=\mathrm{t}\text{ }{\mathrm{e}}^{-\mathrm{t}}\text{ }\mathrm{dt}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

Integrate both sides of equation (2). It results,

$\begin{array}{c}\text{ }\int \mathrm{x}\text{ }{\mathrm{e}}^{2\mathrm{x}}\text{ }\mathrm{dx}=\int \mathrm{t}\text{ }{\mathrm{e}}^{-\mathrm{t}}\text{ }\mathrm{dt}\\ \frac{1}{2}\int 2\mathrm{x}\text{ }{\mathrm{e}}^{2\mathrm{x}}\text{ }\mathrm{dx}=\left(-1\right)\int -\mathrm{t}\text{ }{\mathrm{e}}^{-\mathrm{t}}\text{ }\mathrm{dt}\end{array}$

$\begin{array}{c}\frac{1}{2}\int \mathrm{x}\text{ }\mathrm{d}\left({\mathrm{e}}^{2\mathrm{x}}\right)=\left(-1\right)\int \mathrm{t}\text{ }\mathrm{d}\left({\mathrm{e}}^{-\mathrm{t}}\right)\\ \frac{1}{2}\left[\mathrm{x}\text{ }{\mathrm{e}}^{2\mathrm{x}}-\int {\mathrm{e}}^{2\mathrm{x}}\text{ }\mathrm{dx}\right]=\left(-1\right)\left[\mathrm{t}\text{ }{\mathrm{e}}^{-\mathrm{t}}-\int {\mathrm{e}}^{-\mathrm{t}}\text{ }\mathrm{dt}\right]\left[\int \mathrm{u}\text{ }\mathrm{dv}=\mathrm{uv}-\int \mathrm{v}\text{ }\mathrm{du}\right]\\ \frac{1}{2}\left[\mathrm{x}\text{ }{\mathrm{e}}^{2\mathrm{x}}-\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right]=\left(-1\right)\left[\mathrm{t}\text{ }{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}\right]+\mathrm{k}\left[\mathrm{k}=\mathrm{IntegratingConstant}\right]\\ {\mathrm{e}}^{2\mathrm{x}}\left(2\mathrm{x}-1\right)=-4\text{ }{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)+4\mathrm{k}\\ {\mathrm{e}}^{2\mathrm{x}}\left(2\mathrm{x}-1\right)+4{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)=\mathrm{C}\left[\mathrm{C}=4\mathrm{k}=\mathrm{Constant}\right]\end{array}$

Therefore, the solution of the given equation is ${{\mathbf{e}}}^{2x}\left(2x-1\right){\mathbf{+}}{\mathbf{4}}{{\mathbf{e}}}^{-t}\left(t+1\right){\mathbf{=}}{\mathbf{C}}$.