(y')^2 - 2xy' + (x-3)^2 = 0

Find the values of x whiches gives real solution to the D.E

I don't even know what the question is asking for..

Question

(y')^2 - 2xy' + (x-3)^2 = 0

Find the values of x whiches gives real solution to the D.E

I don't even know what the question is asking for..

holsteremission · Answer

You might be able to solve for $y'$ by completing the square/quadratic formula. Then the solution will probably involve some root term that has no imaginary component for some range of $x$. I think it's this range that you're looking for.

trojanpoem · Answer

Let's try

trojanpoem · Answer

(y' - x)^2 - x^2 + (x-3)^2 = 0

(y'-x)^2 = x^2 - (x-3)^2

(y'-x)^2 = x^2 - (x^2 + 9 - 6x)
   = x^2 - x^2 - 9 + 6x

(y'-x)^2 = 6x - 9

y'-x = sqrt(6x - 9)


6x- 9 >= 0

6x >= 9

x >= 3/2

trojanpoem · Answer

Respect bro,

holsteremission · Answer

Looks good to me. Happy to help :)

trojanpoem · Answer

Thx thx ^^