Solution to Differential Equation/Examples/Arbitrary Order 2 Degree 3 ODE
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Examples of Solutions to Differential Equations
Consider the equation:
- $(1): \quad y = x^2$
where $x \in \R$.
Then $(1)$ is a solution to the second order ODE:
- $(2): \quad \paren {y''}^3 + \paren {y'}^2 - y - 3 x^2 - 8 = 0$
defined on the domain $x \in \R$.
Proof
\(\ds y\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y'\) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y''\) | \(=\) | \(\ds 2\) | Power Rule for Derivatives |
Then:
\(\ds \) | \(\) | \(\ds \paren {2}^3 + \paren {2 x}^2 - x^2 - 3 x^2 - 8\) | substituting for $y$, $y'$ and $y''$ from above into the left hand side of $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 + 4 x^2 - x^2 - 3 x^2 - 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | which equals the right hand side of $(1)$ |
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Example $3.5$