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

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

- 1963: Morris Tenenbaum and Harry Pollard:
*Ordinary Differential Equations*... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Comment $3.43$