First Order ODE/(x + y) dx = (x - y) dy/Proof 2

Theorem

is a homogeneous differential equation with general solution:

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

Proof

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

\(\ds \leadsto \ \ \)

\(\ds \arctan \dfrac y x\)

\(=\)

\(\ds \ln \sqrt {x^2 + y^2} + C\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Problem $4 \ \text{(d)}$