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

Theorem

The second order ODE:

$y y + \paren {y'}^2 - 2 y y' = 0$

has the general solution:

$y^2 = C_2 e^{2 x} + C_1$

Proof

Using Solution of Second Order Differential Equation with Missing Independent Variable:

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

\(=\)

\(\ds 0\)

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

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds p y\)

\(=\)

\(\ds y^2 + C\)

Linear First Order ODE: $y' + \dfrac y x = k x^n$: $k = 2, n = 0$

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \frac {y^2 + C} y\)

\(\ds \leadsto \ \ \)

\(\ds \int \dfrac {y \rd y} {y^2 + C}\)

\(=\)

\(\ds \int \d x\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds \frac 1 2 \, \map \ln {y^2 + C}\)

\(=\)

\(\ds x + k\)

Primitive of Function under its Derivative

After algebra, and reassigning constants:

$y^2 = C_2 e^{2 x} + C_1$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $7$