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 101: 25

Answer

An integrating factor is $\mu(x)= e^{3x} $ So The differential equation : $$ \left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0 $$ becomes exact when multiplied by the integrating factor $\mu(x)= e^{3x} $ and its solutions are given implicitly by $$ e^{3 x} (3yx^{2}+y^{3}) =c $$ where $ c $ is an arbitrary constant

Work Step by Step

$$ \left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0 \quad \quad (i) $$ Comparing this Equation with the differential form: $$ M(x,y) d x+N(x,y) d y=0 $$ we observe that $$ \begin{aligned}M(x,y) &=\left(3 x^{2} y+2 x y+y^{3}\right) \\ N(x,y) &=\left(x^{2}+y^{2}\right) \end{aligned} $$ By calculating $M_{y}$ and $N_{x}$ , we find that $$ \begin{aligned} M_{y}(x, y) &= 3 x^{2}+2 x +3 y^{2} \\ N_{x}(x, y) & = 2x \end{aligned} $$ we find that $$ \frac{M_{y}(x, y)-N_{x}(x, y)}{N(x, y)}=\frac{\left(3 x^{2}+2 x +3 y^{2} \right) -2x }{\left(x^{2}+y^{2}\right) }=3. $$ Thus there is an integrating factor $μ $ that is a function of $x$ only, and it satisfies the differential equation $$ \frac{d \mu}{d x}=3 \mu . $$ Hence $$ \frac{d \mu}{ \mu}=3 dx $$ Integrating, we obtain $$ \mu=e^{3x} $$ Multiplying Eq. (i) by this integrating factor, we obtain $$ \left(3 x^{2} y+2 x y+y^{3}\right) e^{3x} d x+\left(x^{2}+y^{2}\right)e^{3x} d y=0 $$ The latter equation is exact, since $$ M_{y}(x, y)=\left(3 x^{2} +2 x+3y^{2 }\right) e^{3x} =N_{x}(x, y). $$ Thus there is a $\psi (x, y)$ such that $$ \begin{aligned} \psi_{x}(x, y) &=M(x,y) =\left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} , \\ \psi_{y}(x, y) &=N(x,y) =\left(x^{2}+y^{2} \right) e^{3x} \end{aligned} \quad \quad (ii) $$ Integrating the first of these equations with respect to $x$ , we obtain $$ \begin{split} \psi(x, y) & =\int { \left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} } dx \\ & = 3y \int { x^{2} e^{3x}dx}+2y \int {x e^{3x}dx}+y^{3} \int { e^{3x}dx } +h(y) \\ & = 3y (\frac{1}{3} e^{3 x} x^{2}-\frac{2}{27}\left(3 e^{3 x} x-e^{3 x}\right)) +2y (\frac{1}{9}\left(e^{3 x} \cdot 3 x-e^{3 x}\right)) +y^{3} (\frac{1}{3} e^{3 x}) +h(y) \\ & = e^{3 x} (yx^{2}+\frac{1}{3}y^{3})+h(y) \quad \quad \quad\quad\quad (iii) \end{split} $$ where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining $$ \psi_{y}(x, y)= e^{3 x} (x^{2}+y^{2}) +h^{\prime}(y)= \left(x^{2}+y^{2} \right) e^{3x} $$ Thus $h^{\prime}(y)=0 $ and $h(y)=0$ , the constant of integration can be omitted since any solution of the preceding differential equation is satisfactory. Now substituting for $h(y)$ in Eq. (iii) gives $$ \psi(x, y)= e^{3 x} (yx^{2}+\frac{1}{3}y^{3}) $$ Hence solutions of Eq. (i) are given implicitly by $$ \psi(x, y) = e^{3 x} (yx^{2}+\frac{1}{3}y^{3}) =c_{1} $$ where $ c_{1} $ is an arbitrary constant. The solution can be rewritten as follows $$ \psi(x, y) = e^{3 x} (3yx^{2}+y^{3}) =c $$ where $ c=3c_{1} $ is an arbitrary constant
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.