Question

The following illustrate the fact that the superposition principle does not generally hold for nonlinear equations. Will post more info on the question.

Answer 1

Show that \[y = \frac{ 1 }{ x }\] is a solution of \[y' + y^2 =0\] but that if \[c \neq 0\] and \[c \neq 1\]then \[y = \frac{ c }{ x }\] is not a solution. What I have done so far is show that y = 1/x is a solution to the DEQ but I am getting hungup on the but that if part of the problem any help on this would be great :)

Answer 2

Ok, say we have:
$$
y = \frac{c}{x} = c \cdot x^{-1} \\
y' =\left( c \cdot x^{-1} \right)' = c \cdot \left(x^{-1} \right)' = c \cdot \left(-1 \cdot x^{-2} \right) = -c \cdot x^{-2} \\
y' + y^2 = 0 \\
\left( -c \cdot x^{-2} \right) + \left( c \cdot x^{-1} \right)^2 = 0 \\
-c \cdot x^{-2} + c^2 \cdot x^{-2} = 0 \\
(c^2 - c) \cdot x^{-2} = 0
$$But since \(x^{-2}\) cannot be equal to 0 then the only case this is true is when:
$$
c^2 - c = 0 \implies c^2 = c
$$Which works for \(c = 0\) and for \(c \neq 0\) we can divide by c:
$$
c^2 = c \implies \frac{c^2}{c} = \frac{c}{c} \implies c = 1
$$So it can only work for \(c=0\) or \(c = 1\)

Answer 3

thank you very much

Answer 4

Sure, you're welcome =)

Answer 5

Please close this post if you're done.
Thanks! :D