First Order ODE/(x + y) dx = (x - y) dy/Proof 1

From ProofWiki
Jump to navigation Jump to search

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 solution:

$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$


Proof

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:

$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=First_Order_ODE/(x_%2B_y)_dx_%3D_(x_-_y)_dy/Proof_1&oldid=601010"