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.

$\blacksquare$