H Hoda New member Joined Feb 25, 2020 Messages 1 Feb 25, 2020 #1 Please, can you help in solving this problem as i solved it by substitution but no use. it is; (x+y)+(x-y)y' = 0 could you integrate it?? please.
Please, can you help in solving this problem as i solved it by substitution but no use. it is; (x+y)+(x-y)y' = 0 could you integrate it?? please.
D Deleted member 4993 Guest Feb 25, 2020 #2 Hoda said: Please, can you help in solving this problem as i solved it by substitution but no use. it is; (x+y)+(x-y)y' = 0 could you integrate it?? please. Click to expand... Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this assignment.
Hoda said: Please, can you help in solving this problem as i solved it by substitution but no use. it is; (x+y)+(x-y)y' = 0 could you integrate it?? please. Click to expand... Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this assignment.
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Feb 25, 2020 #3 I would arrange the ODE as: [MATH]\d{y}{x}=\frac{y+x}{y-x}=\frac{\dfrac{y}{x}+1}{\dfrac{y}{x}-1}[/MATH] Now, use the substitution: [MATH]v=\frac{y}{x}\implies \d{y}{x}=v+x\d{v}{x}[/MATH]
I would arrange the ODE as: [MATH]\d{y}{x}=\frac{y+x}{y-x}=\frac{\dfrac{y}{x}+1}{\dfrac{y}{x}-1}[/MATH] Now, use the substitution: [MATH]v=\frac{y}{x}\implies \d{y}{x}=v+x\d{v}{x}[/MATH]