First Order ODE/x y' = Root of (x^2 + y^2)

Theorem

The first order ordinary differential equation:

$(1): \quad x y' = \sqrt {x^2 + y^2}$

is a homogeneous differential equation with solution:

$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$

Proof

We divide through by $x$ to show that $(1)$ is homogeneous:

\(\ds \frac {\d x} {\d y}\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \sqrt {1 + \paren {\frac y x}^2}\)

By Solution to Homogeneous Differential Equation, its solution is:

$\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:

$\map f {x, y} = \dfrac {\sqrt {x^2 + y^2} } x$

Thus:

\(\ds \ln x\)

\(=\)

\(\ds \int \frac {\d z} {\sqrt {1 + z^2} - z}\)

\(\ds \)

\(=\)

\(\ds \int \paren {\sqrt{1 + z^2} + z} \rd z\)

\(\ds \)

\(=\)

\(\ds \frac {z \sqrt {1 + z^2} } 2 + \frac {\map \ln {\sqrt{1 + z^2} + z} } 2 + \frac {z^2} 2 + k\)

Primitive of $\sqrt {x^2 + a^2}$

\(\ds \leadsto \ \ \)

\(\ds 2 \ln x\)

\(=\)

\(\ds z \sqrt {1 + z^2} + \map \ln {\sqrt{1 + z^2} + z} + z^2 + C\)

where $C = 2 k$

\(\ds \)

\(=\)

\(\ds \frac y x \sqrt {1 + \paren {\frac y x}^2} + \map \ln {\sqrt {1 + \paren {\frac y x}^2} + \frac y x} + \paren {\frac y x}^2 + C\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds 2 x^2 \ln x\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds 3 x^2 \ln x\)

\(=\)

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

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $4$