Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0
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Theorem
The second order ODE:
- $(1): \quad \paren {x^2 + 2 y'} y + 2 x y' = 0$
subject to the initial conditions:
- $y = 1$ and $y' = 0$ when $x = 0$
has the particular solution:
- $y = 1$
or:
- $3 y + x^3 = 3$
Proof
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.
Substitute $p$ for $y'$ in $(1)$:
\(\ds \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 x p \rd x + \paren {x^2 + 2 p} \rd p\) | \(=\) | \(\ds 0\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds p \paren {x^2 + p}\) | \(=\) | \(\ds C_1\) | Bernoulli's Equation: $2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$ |
Consider the initial condition:
- $y' = p = 0$ when $x = 0$
Hence putting $p = x = 0$ in $(2)$ we get:
- $0 \cdot 0^2 + 0^2 = C_1$
- $C_1 = 0$
and so $(2)$ becomes:
\(\ds p x^2\) | \(=\) | \(\ds -p^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds p \paren {x^2 - p}\) | \(=\) | \(\ds 0\) |
There are two possibilities here:
\(\ds p\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d y} {\d x}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds C_2\) |
From our initial condition:
- $y = 1$ when $x = 0$
gives us:
- $C_2 = 1$
and so the solution is obtained:
- $y = 1$
$\Box$
The other option is:
\(\ds p = \dfrac {\d y} {\d x}\) | \(=\) | \(\ds -x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds -\int x^2 \rd x\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds -\frac {x^3} 3 + C_2\) |
From our initial condition:
- $y = 1$ when $x = 0$
Hence putting $x = 0$ and $y = 1$ in $(3)$ we get:
- $1 = - \dfrac {0^3} 3 = C_2$
and so $C_2 = 1$.
Thus we have:
- $y + \dfrac {x^3} 3 = 1$
or:
- $3 y + x^3 = 3$
Hence the result.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $2 \ \text{(a)}$