Cant solve diff-eq with substitution

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Homework Help Overview

The discussion revolves around solving a differential equation of the form (x^2+y^2)+2xy(dy/dx)=0 using the substitution y=xv, where x is the independent variable and y is dependent on x. Participants are exploring the implications of this substitution and its effects on the equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation using the substitution y=xv and has derived a related expression. Some participants question the nature of v, seeking clarification on whether it is a function or a constant. Others suggest substituting dy/dx in terms of v and its derivative to further analyze the equation.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the substitution and its implications. There is a focus on understanding the role of v and how to properly apply the substitution in the context of the differential equation.

Contextual Notes

Participants are working under the constraints of the problem as a homework assignment, which may limit the information available for discussion. The original poster expresses uncertainty about their previous attempts, indicating a need for guidance without providing a complete solution.

Ryoukomaru
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Use subs [tex]y=xv[/tex] to show that [tex](x^2+y^2)+2xy\frac{dy}{dx}=0, x>0[/tex] is [tex]x^3+3xy^2=k[/tex] where k is a constant.

I played around with this at school and if memory serves me correct i got something similar to [tex]\frac{dx}{dv}=\frac{-3}{2xv}-\frac{1}{2}[/tex] and after that i decided i wasnt on the right path and stopped. Need a little help here ! :smile:
 
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x is to remain the independent variable; v(x) is the dependent variable which replaces y(x)
 
What is v? Is v a function or a constant??
 
stunner5000pt said:
What is v? Is v a function or a constant??
v is defined as the function:
[tex]v(x)=\frac{y(x)}{x}[/tex]
 
If y= xv, then dy/dx= x dv/dx+ v. Put that into your equation and replace y by xv and see what happens.
 

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