Suggested languages for you:

Americas

Europe

Q. 39

Expert-verifiedFound in: Page 570

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{dy}{dx}={e}^{x+y},y\left(0\right)=2$

The solution of the initial-value problem descibed by$\frac{dy}{dx}={e}^{x+y},y\left(0\right)=2$ is $y\left(x\right)=-\mathrm{ln}|1+{e}^{-2}-{e}^{x}|$

The given initial value problem $\frac{dy}{dx}={e}^{x+y},y\left(0\right)=2..............\left(1\right)$

Use the property of exponential function and rewrite the differential equation in (1) as$\frac{dy}{dx}={e}^{x}.{e}^{y}(\because {e}^{x}.{e}^{y}={e}^{x+y})$Note that the differentlal equation is now in the form $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$ In which $p\left(x\right)={e}^{x}$, and $q\left(y\right)={e}^{y}$. So, the differential equation can be solved by applying variable separable method. Thus, the solution of the differential equation involved in the initial- value problem is given by

$\int \frac{1}{{e}^{y}}dy=\int {e}^{x}dx\phantom{\rule{0ex}{0ex}}\int {e}^{-y}dy={e}^{x}+C\phantom{\rule{0ex}{0ex}}-{e}^{-y}={e}^{x}+C\phantom{\rule{0ex}{0ex}}{e}^{-y}=-{e}^{x}-C$

Now, use the given initial condition $y\left(0\right)=2$, that is take $x=0,y=2$ in the above result and evaluate the constant *C.*

${e}^{-2}=-1-C\phantom{\rule{0ex}{0ex}}C=-{e}^{-2}-1$

Substitute this value of the constant *C* in the solution of the differential equation and obtain the solution of the initial - value problem $\frac{dy}{dx}={e}^{x+y},y\left(0\right)=2$ as

${e}^{-y}=-{e}^{x}+1+{e}^{-2}\phantom{\rule{0ex}{0ex}}-y=\mathrm{ln}|1+{e}^{-2}-{e}^{x}|\phantom{\rule{0ex}{0ex}}y=-\mathrm{ln}|1+{e}^{-2}-{e}^{x}|$

Therefore,the solution of the initial-value problem descibed by equation(1) is

$y\left(x\right)=-\mathrm{ln}|1+{e}^{-2}-{e}^{x}|$

94% of StudySmarter users get better grades.

Sign up for free