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