Linear Second Order ODE/y'' - 2 y' + y = 2 x
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 1
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 $(1)$.
Taking the expression $\map R x = 2 x$ and differentiating twice with respect to $x$:
| \(\ds \map R x\) | \(=\) | \(\ds 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {R'} x\) | \(=\) | \(\ds 2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map {R' '} x\) | \(=\) | \(\ds 0\) |
Trying out $y = 2 x$ in $(1)$:
- $0 - 4 + 2 x = 2 x - 4$
which is off by that constant of $4$.
But by Derivative of Constant:
- $4' = 0$
Hence setting $y = 2 x + 4$:
- $0 - 4 + \paren {2 x + 4} = 2 x$
and it can be seen that $y_p = 2 x + 4$ is indeed a particular solution to $(1)$.
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 $(1)$.
$\blacksquare$
Proof 2
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 $(1)$.
$\blacksquare$
Proof 3
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients 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:
- $y_g = C_1 e^x + C_2 x e^x$
We have that:
- $R \left({x}\right) = 2 x$
So from the Method of Undetermined Coefficients for Polynomial:
- $y_p = A_0 + A_1 x$
Hence:
| \(\ds y_p\) | \(=\) | \(\ds A_0 + A_1 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds A_1\) | Power Rule for Derivatives | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds {y_p}\) | \(=\) | \(\ds 0\) | Derivative of Constant |
Substituting into $(1)$:
| \(\ds 0 - 2 A_1 + \paren {A_0 + A_1 x}\) | \(=\) | \(\ds 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds -2 A_1 + A_0\) | \(=\) | \(\ds 0\) | equating coefficients | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A_1\) | \(=\) | \(\ds 2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A_0\) | \(=\) | \(\ds 4\) |
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 $(1)$.
$\blacksquare$