Linear Second Order ODE/y'' - 2 y' + y = 2 x

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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 $(1)$.

$\blacksquare$