A Linear equation of the first order

[chronos280]

I am having some problems with what at first glance seemed to be a fairly simple equation, which is as follows:
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Solve the following equation and write a particular solution for given initial condition (y' means the differentiation with respect to x)

y^2+x^2y'=xyy', y(x=1)=1

Please represent your answer in the implicit form (f(x,y)=0)
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First off I'm unsure of why they have put y' on both sides of the equation, but hey fair enough. My main problem I have is that I don't feel that a solution exists for the condition's given, as one of the first steps I do seems to contradict y(x=1)=1

y^2+x^2y'=xyy'

y^2=(xy-x^2)*y'

y'=(y^2)/(xy-x^2) but this means that (xy-x^2) != 0 but if y=1 when x=1 then this is a contradiction.

In short am I making some obvious or otherwise mistake, or is this question kaput?

 

[Deleted member 4993]

I am having some problems with what at first glance seemed to be a fairly simple equation, which is as follows:
------------------------------------------------------------------------------------------
Solve the following equation and write a particular solution for given initial condition (y' means the differentiation with respect to x)

y^2+x^2y'=xyy', y(x=1)=1

Please represent your answer in the implicit form (f(x,y)=0)
-----------------------------------------------------------------------------------------

First off I'm unsure of why they have put y' on both sides of the equation, but hey fair enough. My main problem I have is that I don't feel that a solution exists for the condition's given, as one of the first steps I do seems to contradict y(x=1)=1

y^2+x^2y'=xyy'

y^2=(xy-x^2)*y'

y'=(y^2)/(xy-x^2) but this means that (xy-x^2) != 0 but if y=1 when x=1 then this is a contradiction

Not really - it only means that at x=1 the the function has a tangent that is vertical.

Divide the equation by x[sup:184nkkvt]2[/sup:184nkkvt]

\(\displaystyle \frac{y^2}{x^2} \ + y' \ = \ y'\frac{y}{x}\)

Now substitute:

\(\displaystyle u = \frac{y}{x}\)

Now continue......

In short am I making some obvious or otherwise mistake, or is this question kaput?

 

[chronos280]

I was getting unnecessarily hung up on the conditions, many thanks for your help.