First Order ODE/(x + y) dx = (x - y) dy
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Theorem
The first order ordinary differential equation:
- $(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$
is a homogeneous differential equation with general solution:
- $\arctan \dfrac y x = \ln \sqrt {x^2 + y^2} + C$
Proof 1
Let:
- $\map M {x, y} = x + y$
- $\map N {x, y} = x - y$
We have that:
- $\map M {t x, t y} = t x + t y = t \paren {x + y} = t \map M {x, y}$
- $\map N {t x, t y} = t x - t y = t \paren {x - y} = t \map N {x, y}$
Thus both $M$ and $N$ are homogeneous functions of degree $1$.
Thus by definition $(1)$ is a homogeneous differential equation.
| \(\ds \frac {\d y} {\d x}\) | \(=\) | \(\ds \frac {x + y} {x - y}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 + y / x} {1 - y / x}\) | dividing top and bottom by $x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {1 + z} {1 - z}\) | substituting $z$ for $y / x$ |
By Solution to Homogeneous Differential Equation:
- $\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
where:
- $\map f {1, z} = \dfrac {1 + z} {1 - z}$
Hence:
| \(\ds \ln x\) | \(=\) | \(\ds \int \frac {\d z} {\dfrac {1 + z} {1 - z} - z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac {\paren {1 - z} \rd z} {1 + z - z \paren {1 - z} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac {\paren {1 - z} \rd z} {1 + z^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \frac {\d z} {1 + z^2} - \int \frac {z \rd z} {1 + z^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \arctan z - \int \frac {z \rd z} {1 + z^2} + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \arctan z - \frac 1 2 \map \ln {1 + z^2} + C\) | Primitive of $\dfrac x {x^2 + a^2}$ |
Substituting $y / x$ for $z$ reveals the solution:
- $\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
$\blacksquare$
Proof 2
We have:
| \(\ds \paren {x + y} \rd x\) | \(=\) | \(\ds \paren {x - y} \rd y\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x \rd y - y \rd x\) | \(=\) | \(\ds x \rd x + y \rd y\) | rearranging | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {x \rd y - y \rd x} {x^2 + y^2}\) | \(=\) | \(\ds \frac {x \rd x + y \rd y} {x^2 + y^2}\) | dividing through by $x^2 + y^2$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {x \rd y - y \rd x} {x^2 + y^2}\) | \(=\) | \(\ds \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }\) | Differential of Sum of Squares | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \d {\arctan \dfrac y x}\) | \(=\) | \(\ds \frac {\map \d {x^2 + y^2} } {2 \paren {x^2 + y^2} }\) | Differential of Arctangent of Quotient | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \arctan \dfrac y x\) | \(=\) | \(\ds \frac {\map \ln {x^2 + y^2} } 2 + C\) | integrating | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \arctan \dfrac y x\) | \(=\) | \(\ds \ln \sqrt {x^2 + y^2} + C\) |
$\blacksquare$