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

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

$1 + x^2 > 0$

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 $(2)$.

$\blacksquare$