Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.9 Exact Differential Equations - Problems - Page 92: 10

Answer

$C=y^2e^{2x}+x^3$

Work Step by Step

We are given: $$(2y^2e^{2x}+3x^2)dx+2ye^{2x}dy=0 (1)$$ Here, $M(x,y)=\frac{\partial}{\partial y}(2y^2e^{2x}+3x^2)=4ye^{2x}$ $N(x,y)=\frac{\partial}{\partial x}(2ye^{2x})=4ye^{2x}$ $M_y=4ye^{2x}\;\;\;\;,\;\;\;\;N_x=4ye^{2x}$ $M_y=N_x$ $\Rightarrow$ (1) is exact differential equation Therefore there exists a potential function $\phi$ such that $\frac{\partial\phi}{\partial x}=2y^2e^{2x}+3x^2$ ___(2) Integrating (2) with respect to $x$ holding $y$ fixed $\phi =y^2e^{2x}+x^3+f(y)$ ___(3) Where $f(y)$ is arbitrary function of $y$ Hence we get: $\frac{\partial\phi}{\partial y}=2y^2e^{2x}+f'(y)=2ye^{2x}$ (4) We have $f'(y)= \rightarrow f(y)=C$ $C$ is constant The solution is: $\phi(x,y)=y^2e^{2x}+x^3+C$ and $\phi(x,y)=0$ so: $C=y^2e^{2x}+x^3$
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