Second Order ODE/y y'' = y^2 y' + (y')^2
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Theorem
The second order ODE:
- $(1): \quad y y = y^2 y' + \paren {y'}^2$
subject to the initial conditions:
- $y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
has the particular solution:
- $2 y - 3 = 8 y \, \map \exp {\dfrac {3 x} 2}$
Proof
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:
| \(\ds y p \frac {\d p} {\d y}\) | \(=\) | \(\ds y^2 p + p^2\) | where $p = \dfrac {\d y} {\d x}$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d p} {\d y} - \frac p y\) | \(=\) | \(\ds y\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds p = \dfrac {\d y} {\d x}\) | \(=\) | \(\ds y \paren {y + C_1}\) | Linear First Order ODE: $y' - \dfrac y x = k x$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \frac {\d y} {y \paren {y + C_1} }\) | \(=\) | \(\ds \int \rd x\) | Solution to Separable Differential Equation | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {C_1} \map \ln {\frac y {y + C_1} }\) | \(=\) | \(\ds x + C_2\) | Primitive of Reciprocal of $\dfrac x {\paren {a x + b} }$ |
Now to consider the initial conditions:
- $y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
After algebra:
| \(\ds \map \ln {\frac y {y + C_1} }\) | \(=\) | \(\ds C_1 x + C_2\) | reassigning $C_2$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac y {y + C_1}\) | \(=\) | \(\ds e^{C_1 x + C_2}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds e^{C_2} e^{C_1 x}\) |
When $x = 0$ we have $y = -1/2$:
| \(\ds \frac {-1/2} {-1/2 + C_1}\) | \(=\) | \(\ds \frac 1 {1 - 2 C_1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds e^{C_2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac y {y + C_1}\) | \(=\) | \(\ds \frac {e^{C_1 x} } {1 - 2 C_1}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \paren {1 - 2 C_1}\) | \(=\) | \(\ds \paren {y + C_1} e^{C_1 x}\) |
Differentiating to get $y'$:
- $y' \paren {1 - 2 C_1} = \paren {y + C_1} C_1 e^{C_1 x} + e^{C_1 x} y'$
Putting $y' = 1$ when $x = 0$ we get:
| \(\ds 1 - 2 C_1\) | \(=\) | \(\ds \paren {-\frac 1 2 + C_1} C_1 + 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C_1\) | \(=\) | \(\ds -\frac 3 2\) |
So:
| \(\ds y \paren {1 - 2 C_1}\) | \(=\) | \(\ds \paren {y + C_1} e^{C_1 x}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \paren {1 - 2 \frac {-3} 2}\) | \(=\) | \(\ds \paren {y - \frac 3 2} e^{\frac {-3 x} 2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 8 y\) | \(=\) | \(\ds \paren {2 y - 3} e^{\frac {-3 x} 2}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 y - 3\) | \(=\) | \(\ds 8 y \, \map \exp {\dfrac {3 x} 2}\) |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $2 \ \text{(b)}$