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

Theorem

The second order ODE:

$x^2 y = 2 x y' + \paren {y'}^2$

has the general solution:

$y = -\dfrac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2$

Proof

The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.

Substitute $p$ for $y'$:

\(\ds x^2 \dfrac {\d p} {\d x}\)

\(=\)

\(\ds 2 x p + p^2\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds -\frac {x^2} {x - C_1}\)

Bernoulli's Equation: $x^2 \rd y = \paren {2 x y + y^2} \rd x$

\(\ds \leadsto \ \ \)

\(\ds \int \rd y\)

\(=\)

\(\ds -\int \frac {x^2} {x - C_1} \rd x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds -\frac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2\)

Primitive of $\dfrac {x^2} {a x + b}$

$\blacksquare$

Sources

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