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Q. 27

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 the given differential equations. Your answers will involve unsolved constants.

$\frac{dy}{dx}=x{e}^{-y}$

Ans: The solution of the differential equation $\frac{dy}{dx}=x{e}^{-y}$ is $y=\mathrm{ln}\left|\frac{1}{2}{x}^{2}+C\right|$

given,

$\frac{dy}{dx}=x{e}^{-y}$

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

Note that the differential equation (1) is of the form of $\frac{dy}{dx}=p\left(x\right)q\left(y\right)$ in which $p\left(x\right)=x$ and $q\left(y\right)={e}^{-y}$. So the differential equation can be solved by applying the variable separable method. Separate the variables and integrate both the sides

$\begin{array}{r}\int \frac{1}{{e}^{-y}}dy=\int xdx\\ \int {e}^{y}dy=\frac{1}{2}{x}^{2}+C\\ {e}^{y}=\frac{1}{2}{x}^{2}+C\\ y=\mathrm{ln}\left|\frac{1}{2}{x}^{2}+C\right|\end{array}$

Hence a solution to the differential equation $\frac{dy}{dx}=x{e}^{-y}$ isrole="math" localid="1649178838912" $y=\mathrm{ln}\left|\frac{1}{2}{x}^{2}+C\right|$

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