Linear Second Order ODE/y'' - 4 y' + 4 y = 0
Theorem
The second order ODE:
- $(1): \quad y - 4 y' + 4 y = 0$
has the general solution:
- $y = C_1 e^{2 x} + C_2 x e^{2 x}$
Proof 1
Consider the functions:
- $\map {y_1} x = e^{2 x}$
- $\map {y_2} x = x e^{2 x}$
We have that:
| \(\ds \frac \d {\d x} \, e^{2 x}\) | \(=\) | \(\ds 2 e^{2 x}\) | Power Rule for Derivatives | |||||||||||
| \(\ds \frac \d {\d x} \, x e^{2 x}\) | \(=\) | \(\ds 2 x e^{2 x} + e^{2 x}\) | Power Rule for Derivatives | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d^2} {\d x^2} \, e^{2 x}\) | \(=\) | \(\ds 4 e^{2 x}\) | |||||||||||
| \(\ds \frac {\d^2} {\d x^2} \, e^{2 x}\) | \(=\) | \(\ds 4 x e^{2 x} + 2 e^{2 x} + 2 e^{2 x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4 x e^{2 x} + 4 e^{2 x}\) |
Putting $e^{2 x}$ and $x e^{2 x}$ into $(1)$ in turn:
| \(\ds 4 e^0 - 4 \times 2 e^0 + 4 e^0\) | \(=\) | \(\ds 8 - 8 + 4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
| \(\ds 4 \times 0 e^0 + 4 e^0 - 4 \paren {2 \times 0 e^0 + e^0} + 4 \paren {0 e^0}\) | \(=\) | \(\ds 0 + 4 - 4 + 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Hence it can be seen that:
| \(\ds \map {y_1} x\) | \(=\) | \(\ds e^{2 x}\) | ||||||||||||
| \(\ds \map {y_2} x\) | \(=\) | \(\ds x e^{2 x}\) |
are particular solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
| \(\ds \map W {y_1, y_2}\) | \(=\) | \(\ds \begin{vmatrix} e^{2 x} & x e^{2 x} \\ 2 e^{2 x} & \paren {2 x + 1} e^{2 x} \end{vmatrix}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds e^{2 x} \times \paren {2 x + 1} e^{2 x} - x e^{2 x} \times 2 e^{2 x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {2 x + 1} e^{4 x} - 2 x e^{4 x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds e^{4 x}\) |
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
- $y_1$ and $y_2$ are linearly independent everywhere on $\R$.
We have that $(1)$ is a homogeneous linear second order ODE in the form:
- $y + \map P x y' + \map Q x y = 0$
where $\map P x = -4$ and $\map Q x = 4$.
So by Constant Real Function is Continuous:
- $P$ and $Q$ are continuous on $\R$.
Thus from Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:
- $(1)$ has the general solution:
- $y = C_1 e^{2 x} + C_2 x e^{2 x}$
$\blacksquare$
Proof 2
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
- $(2): \quad: m^2 - 4 m + 4 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
- $m_1 = m_2 = 2$
These are real and equal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
- $y = C_1 e^{2 x} + C_2 x e^{2 x}$
$\blacksquare$