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### 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

# ${I}{n}{}{p}{r}{o}{b}{l}{e}{m}{}{7}{-}{16}{}{,}{}{s}{o}{l}{v}{e}{}{t}{h}{e}{}{e}{q}{u}{a}{t}{i}{o}{n}{.}{}{}\frac{dx}{dt}{-}{{x}}^{{3}}{=}{x}$

$Thesolutionofthegivendifferentialequationis\frac{\mathbf{x}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{=}}{\mathbit{C}}{{\mathbit{e}}}^{{\mathbf{t}}}{\mathbf{}}{\mathbf{.}}$

See the step by step solution

## Step 1: Concept of Separable Differential Equation

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t(}\mathbf{y}\mathbf{\right)}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{x}}{\mathbf{}}{\mathbf{.}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{T}}{\mathbit{h}}{\mathbit{e}}{\mathbit{n}}{\mathbf{,}}{\mathbf{}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{r}}{\mathbit{e}}{\mathbit{c}}{\mathbit{t}}{\mathbf{}}{\mathbit{i}}{\mathbit{n}}{\mathbit{t}}{\mathbit{e}}{\mathbit{g}}{\mathbit{r}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{b}}{\mathbit{o}}{\mathbit{t}}{\mathbit{h}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{d}}{\mathbit{e}}{\mathbit{s}}{\mathbf{,}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{s}}{\mathbit{o}}{\mathbit{l}}{\mathbit{u}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{f}}{\mathbit{f}}{\mathbit{e}}{\mathbit{r}}{\mathbit{e}}{\mathbit{n}}{\mathbit{t}}{\mathbit{i}}{\mathbit{a}}{\mathbit{l}}{\mathbf{}}{\mathbit{e}}{\mathbit{q}}{\mathbit{u}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{i}}{\mathbit{s}}{\mathbf{}}{\mathbit{d}}{\mathbit{e}}{\mathbit{t}}{\mathbit{e}}{\mathbit{r}}{\mathbit{m}}{\mathbit{i}}{\mathbit{n}}{\mathbit{e}}{\mathbit{d}}{\mathbf{.}}{\mathbf{}}$$Afirst-orderordinarydifferentialequation\frac{dy}{dx}=f\left(x,y\right)isreferredtoasseparableifthefunctionintheright-handsideoftheequationisexpressedasaproductoftwofunctionsg\left(x\right)thatisafunctionofxaloneandh\left(y\right)thatisafunctionofyalone.\phantom{\rule{0ex}{0ex}}{\mathbit{A}}{\mathbf{}}{\mathbit{s}}{\mathbit{e}}{\mathbit{p}}{\mathbit{a}}{\mathbit{r}}{\mathbit{a}}{\mathbit{b}}{\mathbit{l}}{\mathbit{e}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{f}}{\mathbit{f}}{\mathbit{e}}{\mathbit{r}}{\mathbit{e}}{\mathbit{n}}{\mathbit{t}}{\mathbit{i}}{\mathbit{a}}{\mathbit{l}}{\mathbf{}}{\mathbit{e}}{\mathbit{q}}{\mathbit{u}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{c}}{\mathbit{a}}{\mathbit{n}}{\mathbf{}}{\mathbit{b}}{\mathbit{e}}{\mathbf{}}{\mathbit{e}}{\mathbit{x}}{\mathbit{p}}{\mathbit{r}}{\mathbit{e}}{\mathbit{s}}{\mathbit{s}}{\mathbit{e}}{\mathbit{d}}{\mathbf{}}{\mathbit{a}}{\mathbit{s}}{\mathbf{}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbit{h}}{\mathbf{\left(}}{\mathbit{y}}{\mathbf{\right)}}{\mathbf{.}}{\mathbf{}}{\mathbit{B}}{\mathbit{y}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{s}}{\mathbit{e}}{\mathbit{p}}{\mathbit{a}}{\mathbit{r}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{n}}{\mathbit{g}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{v}}{\mathbit{a}}{\mathbit{r}}{\mathbit{i}}{\mathbit{a}}{\mathbit{b}}{\mathbit{l}}{\mathbit{e}}{\mathbit{s}}{\mathbf{,}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{e}}{\mathbit{q}}{\mathbit{u}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{f}}{\mathbit{o}}{\mathbit{l}}{\mathbit{l}}{\mathbit{o}}{\mathbit{w}}{\mathbit{s}}{\mathbf{}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{h}\mathbf{\left(}\mathbf{y}\mathbf{\right)}}{\mathbf{=}}{\mathbit{g}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbit{d}}{\mathbit{x}}{\mathbf{}}{\mathbf{.}}{\mathbf{}}{\mathbit{T}}{\mathbit{h}}{\mathbit{e}}{\mathbit{n}}{\mathbf{,}}{\mathbf{}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{r}}{\mathbit{e}}{\mathbit{c}}{\mathbit{t}}{\mathbf{}}\phantom{\rule{0ex}{0ex}}{\mathbit{i}}{\mathbit{n}}{\mathbit{t}}{\mathbit{e}}{\mathbit{g}}{\mathbit{r}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{b}}{\mathbit{o}}{\mathbit{t}}{\mathbit{h}}{\mathbf{}}{\mathbit{s}}{\mathbit{i}}{\mathbit{d}}{\mathbit{e}}{\mathbit{s}}{\mathbf{,}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{s}}{\mathbit{o}}{\mathbit{l}}{\mathbit{u}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{o}}{\mathbit{f}}{\mathbf{}}{\mathbit{t}}{\mathbit{h}}{\mathbit{e}}{\mathbf{}}{\mathbit{d}}{\mathbit{i}}{\mathbit{f}}{\mathbit{f}}{\mathbit{e}}{\mathbit{r}}{\mathbit{e}}{\mathbit{n}}{\mathbit{t}}{\mathbit{i}}{\mathbit{a}}{\mathbit{l}}{\mathbf{}}{\mathbit{e}}{\mathbit{q}}{\mathbit{u}}{\mathbit{a}}{\mathbit{t}}{\mathbit{i}}{\mathbit{o}}{\mathbit{n}}{\mathbf{}}{\mathbit{i}}{\mathbit{s}}{\mathbf{}}{\mathbit{d}}{\mathbit{e}}{\mathbit{t}}{\mathbit{e}}{\mathbit{r}}{\mathbit{m}}{\mathbit{i}}{\mathbit{n}}{\mathbit{e}}{\mathbit{d}}{\mathbf{.}}{\mathbf{}}$uncaught 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A first-order ordinary differential equation $\frac{dy}{dx}{=}{f}{\left(}{x}{,}{y}{\right)}$ is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions g(x) that is a function of x alone and h(y) that is a function of y alone.

A separable differential equation can be expressed as $\frac{dy}{dx}{=}{g}{\left(}{x}{\right)}{h}{\left(}{y}{\right)}$ . By separating the variables, the equation follows $\frac{dy}{h\left(y\right)}{=}{g}{\left(}{x}{\right)}{d}{x}$ . 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

$\frac{dx}{dt}-{x}^{3}=x......\left(1\right)$

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

$\frac{dx}{dt}=x+{x}^{3}\phantom{\rule{0ex}{0ex}}=x\left(1+{x}^{2}\right)\phantom{\rule{0ex}{0ex}}\frac{dx}{x\left(1+{x}^{2}\right)}=dt.....\left(2\right)$

Before proceeding for integration, let us decompose the integrand $\frac{1}{x\left(1+{x}^{2}\right)}$ into partial fraction decomposition. It follows,

$\frac{1}{x\left(1+{x}^{2}\right)}=\frac{1}{x}-\frac{x}{1+{x}^{2}}$

Then, equation (2) becomes,

$\left(\frac{1}{x}-\frac{x}{1+{x}^{2}}\right)dx=dt.....\left(3\right)$

Now, integrate both sides of equation (3). It results,

$\int \left(\frac{1}{x}-\frac{x}{1+{x}^{2}}\right)dx=\int dt\phantom{\rule{0ex}{0ex}}\int \frac{dx}{x}-\int \frac{xdx}{1+{x}^{2}}=\int dt\phantom{\rule{0ex}{0ex}}\int \frac{dx}{x}-\frac{1}{2}\int \frac{2xdx}{1+{x}^{2}}=\int dt\phantom{\rule{0ex}{0ex}}\int \frac{dx}{x}-\int \frac{d\left(1+{x}^{2}\right)}{1+{x}^{2}}=\int dt\phantom{\rule{0ex}{0ex}}\mathrm{ln}x-\mathrm{ln}\left(1+{x}^{2}\right)=t+\mathrm{ln}C\left[\mathrm{ln}C=\mathrm{ln}integratingCons\mathrm{tan}t\right]\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\frac{x}{1+{x}^{2}}\right)-\mathrm{ln}C=t\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\frac{x}{C\left(1+{x}^{2}\right)}\right)=t\phantom{\rule{0ex}{0ex}}\frac{x}{1+{x}^{2}}=C{e}^{t}\phantom{\rule{0ex}{0ex}}$ Therefore, the solution of the Therefore, the solution of given equation is $\frac{x}{1+{x}^{2}}{=}{C}{{e}}^{{t}}$ .