Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 2
Theorem
The second order ODE:
- $(1): \quad y - 2 y' + y = 2 x$
has the general solution:
- $y = C_1 e^x + C_2 x e^x + 2 x + 4$
Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
- $y + p y' + q y = \map R x$
where:
- $p = -2$
- $q = 1$
- $\map R x = 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $(2): \quad y - 2 y' + y = 0$
From Linear Second Order ODE: $y - 2 y' + y = 0$, this has the general solution:
- $y_g = C_1 e^x + C_2 x e^x$
It remains to find a particular solution $y_p$ to $(1)$.
Expressing $y_g$ in the form:
- $y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
we have:
\(\ds \map {y_1} x\) | \(=\) | \(\ds e^x\) | ||||||||||||
\(\ds \map {y_2} x\) | \(=\) | \(\ds x e^x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map { {y_1}'} x\) | \(=\) | \(\ds e^x\) | Derivative of Exponential Function | ||||||||||
\(\ds \map { {y_2}'} x\) | \(=\) | \(\ds x e^x + e^x\) | Derivative of Exponential Function, Product Rule for Derivatives |
By the Method of Variation of Parameters, we have that:
- $y_p = v_1 y_1 + v_2 y_2$
where:
\(\ds v_1\) | \(=\) | \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\) | ||||||||||||
\(\ds v_2\) | \(=\) | \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\) |
where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.
We have that:
\(\ds \map W {y_1, y_2}\) | \(=\) | \(\ds y_1 {y_2}' - y_2 {y_1}'\) | Definition of Wronskian | |||||||||||
\(\ds \) | \(=\) | \(\ds e^x \paren {x e^x + e^x} - x e^x e^x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x e^{2 x} + e^{2 x} - x e^{2 x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{2 x}\) |
Hence:
\(\ds v_1\) | \(=\) | \(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int -\frac {x e^x 2 x} {e^{2 x} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int -2 x^2 e^{- x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 e^{-x} \paren {x^2 + 2 x + 2}\) | Primitive of $x^2 e^{a x}$ |
\(\ds v_2\) | \(=\) | \(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {e^x 2 x} {e^{2 x} } \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int 2 x e^{-x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 e^{- x} \paren {x + 1}\) | Primitive of $x e^{a x}$ |
It follows that:
\(\ds y_p\) | \(=\) | \(\ds 2 e^{-x} \paren {x^2 + 2 x + 2} e^x - 2 e^{-x} \paren {x + 1} x e^x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 x^2 + 4 x + 4 - 2 x^2 - 2 x\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 x + 4\) | further simplifying |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$
is the general solution to $(1)$.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $1$