Solution to Differential Equation/Examples/Equation which is Not a Solution

From ProofWiki
Jump to navigation Jump to search

Examples of Solutions to Differential Equations

Consider the equation:

$(1): \quad y = \sqrt {-\paren {1 + x^2} }$

where $x \in \R$.

Consider the first order ODE:

$(2): \quad x + y y' = 0$

Then despite the fact that the formal substition for $y$ and $y'$ from $(1)$ into $(2)$ yields an identity, $(1)$ is not a solution to $(2)$.


Proof

First we note that:

\(\ds y\) \(=\) \(\ds \sqrt {-\paren {1 + x^2} }\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds -\dfrac 1 2 \paren {1 + x^2}^{-1/2} {2 x}\) Power Rule for Derivatives, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds -\dfrac x {\sqrt {-\paren {1 + x^2} } }\) Power Rule for Derivatives, Derivative of Constant


Then:

\(\ds \) \(\) \(\ds x + \sqrt {-\paren {1 + x^2} } \paren {-\dfrac x {\sqrt {-\paren {1 + x^2} } } }\) substituting for $y$ and $y'$ from above into the left hand side of $(2)$
\(\ds \) \(=\) \(\ds x - x\)
\(\ds \) \(=\) \(\ds 0\) which equals the right hand side of $(1)$


However, by Domain of Real Square Root Function, $\sqrt {-\paren {1 + x^2} }$ is defined for $-\paren {1 + x^2} \ge 0$.

But by Square of Real Number is Non-Negative:

$1 + x^2 > 0$

and so:

$-\paren {1 + x^2} < 0$

So there exists no $x \in \R$ for which $y = \sqrt {-\paren {1 + x^2} }$ is defined.

So $(1)$ does not define a real function and so $(1)$ is not a solution to $(2)$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Solution_to_Differential_Equation/Examples/Equation_which_is_Not_a_Solution&oldid=624499"