What is a solution to the differential equation #dy/dx=x/(x+y)#?

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Answer 1 · 2017-01-31T12:04:40

See below.

Making #y = lambda x# and knowing that

#dy=lambda dx + x d lambda->(dy)/(dx)=lambda + x(d lambda)/(dx)#

we get at

#lambda + x (d lambda)/(dx)=1/(lambda+1)# or

#x (d lambda)/(dx)=-(lambda^2+lambda-1)/(lambda+1)# and now this differential equation is separable

#(dx)/x=-((lambda+1)d lambda)/(lambda^2+lambda-1)#

so

#(dx)/x = -((lambda+1)d lambda)/((lambda+1/2)^2-5/4)#

giving

#logx=-1/10((sqrt(5)+5)log(sqrt5-1-2lambda)-(sqrt(5)-5)log(sqrt5+1+2lambda))+C#

Now substituting,

#logx=-1/10((sqrt(5)+5)log(sqrt5-1-2y/x)-(sqrt(5)-5)log(sqrt5+1+2y/x))+C#

so the solution appears in implicit form

#G(y(x),x,C)=0#