First Order ODE/x y' = 2 y
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Theorem
The first order ODE:
- $x y' = 2 y$
has the general solution:
- $y = C x^2$
where $C$ is an arbitrary constant.
Proof
\(\ds x y'\) | \(=\) | \(\ds 2 y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} y\) | \(=\) | \(\ds 2 \int \dfrac {\d x} x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds 2 \ln x + \ln C\) | Primitive of Reciprocal | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds \map \ln C x^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds C x^2\) |
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.15)$