StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q2.2-15E

Expert-verifiedFound in: Page 46

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**In problem ****, solve the equation. ${(}{x}{+}{x}{{y}}^{{2}}{)}{d}{x}{+}{{e}}^{{x}^{2}}{y}{}{d}{y}{=}{0}$**

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

**A first-order ordinary differential equation $\frac{dy}{dx}{=}{f}{(}{x}{,}{y}{)}$ 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. **

The given equation is

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

Re-write equation (1) as follows:

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

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,

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

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

94% of StudySmarter users get better grades.

Sign up for free