Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 1

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 $(1)$.

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 $(1)$.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $1$