Second Order ODE/y y'' + (y')^2 = 0

Theorem

The second order ODE:

$(1): \quad y y + \paren {y'}^2 = 0$

has the general solution:

$y^2 = C_1 x + C_2$

Proof

Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

\(\ds y p \frac {\d p} {\d y} + p^2\)

\(=\)

\(\ds 0\)

where $p = \dfrac {\d y} {\d x}$

\(\ds \leadsto \ \ \)

\(\ds y \rd p + p \rd y\)

\(=\)

\(\ds 0\)

multiplying by $\dfrac {\d y} p$

\(\ds \leadsto \ \ \)

\(\ds p y\)

\(=\)

\(\ds C\)

First Order ODE: $y \rd x + x \rd y = 0$

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds C\)

substituting $p = \dfrac {\d y} {\d x}$

\(\ds \leadsto \ \ \)

\(\ds y^2\)

\(=\)

\(\ds 2 C x + C_2\)

First Order ODE: $y \dfrac {\d y} {\d x} = k$

\(\ds \)

\(=\)

\(\ds C_1 x + C_2\)

reassigning constant

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $1 \ \text{(a)}$