First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0
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Theorem
The first order ordinary differential equation:
- $(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
is an exact differential equation with solution:
- $\dfrac {y^2} 2 - \dfrac y x = C$
Proof
Let $M$ and $N$ be defined as:
- $\map M {x, y} = \dfrac y {x^2}$
- $\map N {x, y} = y - \dfrac 1 x$
Then:
\(\ds \frac {\partial M} {\partial y}\) | \(=\) | \(\ds \frac 1 {x^2}\) | ||||||||||||
\(\ds \dfrac {\partial N} {\partial x}\) | \(=\) | \(\ds \dfrac 1 {x^2}\) |
Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.
By Solution to Exact Differential Equation, the solution to $(1)$ is:
- $\map f {x, y} = C$
where:
\(\ds \dfrac {\partial f} {\partial x}\) | \(=\) | \(\ds \map M {x, y}\) | ||||||||||||
\(\ds \dfrac {\partial f} {\partial y}\) | \(=\) | \(\ds \map N {x, y}\) |
Hence:
\(\ds f\) | \(=\) | \(\ds \map M {x, y} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \dfrac y {x^2} \rd x + \map g y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - \frac y x + \map g y\) |
and:
\(\ds f\) | \(=\) | \(\ds \int \map N {x, y} \rd y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {y - \dfrac 1 x} \r d y + \map h x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {y^2} 2 - \frac y x + \map h x\) |
Thus:
- $\map f {x, y} = \dfrac {y^2} 2 - \dfrac y x$
By Solution to Exact Differential Equation, the solution to $(1)$ is:
- $\dfrac {y^2} 2 - \dfrac y x = C$
$\blacksquare$