Can someone please help me answer this question please 5. Solve this Homogeneous First-Order Diff. Equation:Hint: Partial fractions may be required for this problem.(y² + yx) dx + x²dy = 0

given (y2+yx)dx+x2dy=0____(1)⇒dydx=-(y2+yx)x2put y=vx⇒dydx=v+x.dvdxNow the given equation…

Answered: Solve this Homogeneous First-Order…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Can someone please help me answer this question please

5. Solve this Homogeneous First-Order Diff. Equation:
Hint: Partial fractions may be required for this problem.
(y² + yx) dx + x²dy = 0

Expert Solution

Step 1

$g i v e n (y^{2} + y x) d x + x^{2} d y = 0____(1) \Rightarrow \frac{d y}{d x} = - \frac{(y^{2} + y x)}{x^{2}} p u t y = v x \Rightarrow \frac{d y}{d x} = v + x . \frac{d v}{d x} N o w t h e g i v e n e q u a t i o n b e c o m e s v + x . \frac{d v}{d x} = - \frac{{(v x)}^{2} + (v x) x}{x^{2}} = - \frac{x^{2} (v^{2} + v)}{x^{2}} = - v^{2} - v \Rightarrow x . \frac{d v}{d x} = - v^{2} - 2 v = - v (v + 2) \Rightarrow \frac{d v}{v (v + 2)} = - \frac{d x}{x}, w h i c h i s v a r i a b l e s e p a r a b l e f o r m \Rightarrow \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] d v = - \frac{d x}{x} o n i n t e g r a t i o n b o t h s i d e s \int \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] d v = - \int \frac{d x}{x} + c \Rightarrow \frac{1}{2} [\ln (v) - \ln (v + 2)] = - \ln (x) + c \Rightarrow \frac{1}{2} \ln [\frac{v}{v + 2}] = - \ln (x) + c p u t v = \frac{y}{x} \Rightarrow \frac{1}{2} \ln [\frac{(\frac{y}{x})}{(\frac{y}{x}) + 2}] = - \ln (x) + c \Rightarrow \frac{1}{2} \ln [\frac{y}{y + 2 x}] = - \ln (x) + c \Rightarrow \ln {[\frac{y}{y + 2 x}]}^{1 / 2} + \ln (x) = c \Rightarrow \ln [{[\frac{y}{y + 2 x}]}^{1 / 2} . (x)] = c$