Second Order ODE/x^2 y'' + x y' = 1

Theorem

The second order ODE:

$x^2 y + x y' = 1$

has the general solution:

$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$

Proof

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

Substitute $p$ for $y'$:

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

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds x \dfrac {\d p} {\d x} + p\)

\(=\)

\(\ds \frac 1 x\)

\(\ds \leadsto \ \ \)

\(\ds x p\)

\(=\)

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

Linear First Order ODE: $x y' + y = \map f x$

\(\ds \)

\(=\)

\(\ds \ln x + C_1\)

Primitive of Reciprocal

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \frac {\ln x} x + \frac {C_1} x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \int \paren {\frac {\ln x} x + \frac {C_1} x} \rd x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \frac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2\)

Primitive of $\dfrac {\ln x} x$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $16$