First Order ODE/y dy = k x dx
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Theorem
Let $k \in \R$ be a real number.
The first order ODE:
- $y \rd y = k x \rd x$
has the general solution:
- $y^2 = k x^2 + C$
Proof
\(\ds y \rd y\) | \(=\) | \(\ds k x \rd x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int y \rd y\) | \(=\) | \(\ds \int k x \rd x\) | Solution to Separable Differential Equation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {y^2} 2\) | \(=\) | \(\ds \frac {k x^2} 2 + C_1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^2\) | \(=\) | \(\ds k x^2 + C\) | reassigning the arbitrary constant. |
$\blacksquare$