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Q2.2-15E

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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 , solve the equation. ${\left(}{x}{+}{x}{{y}}^{{2}}{\right)}{d}{x}{+}{{e}}^{{x}^{2}}{y}{}{d}{y}{=}{0}$

The solution of the given differential equation is . ${y}{=}{±}\sqrt{Cexp\left[exp\left(-{x}^{2}\right)\right]-1}$

See the step by step solution

## Step 1: Concept of Separable Differential Equation

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

$\left(x+x{y}^{2}\right)dx+{e}^{{x}^{2}}ydy=0.....\left(1\right)$

Re-write equation (1) as follows:

$\left(x+x{y}^{2}\right)dx+{e}^{{x}^{2}}ydy=0\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=-\frac{x\left(1+{y}^{2}\right)}{{e}^{{x}^{2}}y}........\left(2\right)$

After separating the variables, equation (2) can be written as

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

Integrate both sides of equation (3). It results,

$\int \frac{y}{1+{y}^{2}}dy=\int -x{e}^{{x}^{2}}dx\phantom{\rule{0ex}{0ex}}\frac{1}{2}\int \frac{2y}{1+{y}^{2}}dy=\frac{1}{2}\int -2x{e}^{{x}^{2}}dx\phantom{\rule{0ex}{0ex}}\frac{1}{2}\int \frac{d\left(1+{y}^{2}\right)}{1+{y}^{2}}=\frac{1}{2}\int 2{e}^{{x}^{2}}d\left(-{x}^{2}\right)\phantom{\rule{0ex}{0ex}}\frac{1}{2}\mathrm{ln}\left(1+{y}^{2}\right)=\frac{1}{2}{e}^{-{x}^{2}}+\mathrm{ln}k\left[\mathrm{ln}k=Integratingcons\mathrm{tan}t\right]\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(1+{y}^{2}\right)={e}^{-{x}^{2}}+\mathrm{ln}C\left[\mathrm{ln}C=2\mathrm{ln}k=cons\mathrm{tan}t\right]\phantom{\rule{0ex}{0ex}}\mathrm{ln}\left(\frac{1+{y}^{2}}{C}\right)={e}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}1+{y}^{2}=C{e}^{{x}^{2}}\phantom{\rule{0ex}{0ex}}y=±\sqrt{C{e}^{{x}^{2}}-1}\phantom{\rule{0ex}{0ex}}y=±\sqrt{Cexp\left[exp\left({x}^{2}\right)\right]-1}$

Therefore, the solution of the given equation is ${y}{=}{±}\sqrt{Cexp\left[exp\left({x}^{2}\right)\right]-1}$ .

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