Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 2 - First Order Differential Equations - 2.6 Exact Equations and Integrating Factors - Problems - Page 100: 21


The differential equation : $$ y d x+\left(2 x-y e^{y}\right) d y=0 $$ is not exact but becomes exact when multiplied by the given integrating factor $\mu(x, y)=y $ and its solutions are given implicitly by $$ xy^{2} -e^{y}\left( y^{2} -2 y+2 \right)=c $$ where $ c $ is an arbitrary constant

Work Step by Step

$$ y d x+\left(2 x-y e^{y}\right) d y=0, \quad \mu(x, y)=y \quad \quad (i) $$ Comparing this Equation with the differential form: $$ M(x,y) d x+N(x,y) d y=0 $$ we observe that $$ \begin{aligned}M(x,y) &= y \\ N(x,y) &=2 x-y e^{y} \end{aligned} $$ By calculating $M_{y}$ and $N_{x}$ , we find that $$ \begin{aligned} M_{y}(x, y) &= 1 \\ N_{x}(x, y) & =2 \end{aligned} $$ we obtain that $$ M_{y}(x, y) \neq N_{x}(x, y) $$ so the given equation is not exact. Multiplying Eq. (i) by this integrating factor $\mu(x, y)=y $, we obtain $$ y^{2} d x+\left(2y x-y^{2} e^{y}\right) d y=0, \quad \quad (ii) $$ The latter equation is exact, since $$ M_{y}(x, y)=2y=N_{x}(x, y) $$ the equation (ii) is exact. Thus there is a $\psi (x, y)$ such that $$ \begin{aligned} \psi_{x}(x, y) &=M(x,y) =y^{2} , \\ \psi_{y}(x, y) &=N(x,y) =\left(2y x-y^{2} e^{y}\right) \end{aligned} \quad \quad (iii) $$ Integrating the first of these equations with respect to $x$ , we obtain $$ \begin{split} \psi(x, y) & = \int { y^{2} } dx \\ & = xy^{2} +h(y) \quad \quad (*) \end{split} $$ where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (iii), we compute $\psi_{y}(x, y) $ from Eq. (*) and set it equal to $N$, obtaining $$ \psi_{y}(x, y)= 2xy+h^{\prime}(y)= \left(2y x-y^{2} e^{y}\right) $$ Thus $h^{\prime}(y) =-y^{2} e^{y} $ Integrating this Equation by parts gives $$ h(y)=- \int y^{2} e^{y} d y=-y^{2} e^{y}+2\left(e^{y} y-e^{y}\right) $$ The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory. Now substituting for $h(y)$ in Eq. (*) gives $$ \psi(x, y)= xy^{2} -e^{y}\left( y^{2} -2 y+2 \right) $$ Hence solutions of Eq. (i) are given implicitly by $$ \psi(x, y) = xy^{2} -e^{y}\left( y^{2} -2 y+2 \right)=c $$ where $ c $ is an arbitrary constant
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