Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 3

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 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 $(1)$.

$\blacksquare$