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 - 1.11 Some Higher-Order Differential Equations - Problems - Page 106: 13

Answer

$x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+k$

Work Step by Step

We are given: $y''+y^{-1}(y')^2=ye^{-y}(y')^3$ Suppose that $u=\frac{dy}{dx}$ Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$ Subsitute to the equation: $u\frac{du}{dx}+\frac{u^2}{y}=ye^{-y}u^3$ Multiplying both sides by $\frac{1}{u}$ $\frac{du}{dy}+\frac{u}{y}=ye^{-y}u^2$ Then continue to multiply both sides by $\frac{1}{u^2}$: $\frac{1}{u^2}\frac{du}{dy}+\frac{1}{uy}=ye^{-y}$ Let $t=\frac{1}{u} \rightarrow \frac{-du}{u^2}=dt$ $-\frac{dt}{dy}+\frac{t}{y}=ye^{-y}$ $\frac{dt}{dy}-\frac{1}{y}t=ye^{-y}$ Integrating factor: $I=e^{-\int \frac{1}{y}dy}=e^{-\ln y}=y^{-1}$ Integrating both sides: $\frac{1}{y}t=e^{-y}+C$ $\frac{1}{u}=ye^{-y}+Cy$ $u=\frac{1}{ye^{-y}+Cy}$ $\frac{dy}{dx}=\frac{1}{ye^{-y}+Cy}$ Integrating the last equation: $x=\int ye^{-y}dy + C\int ydy$ $x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+C_2$ The final solution is: $x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+k$
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