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

Theorem

The second order ODE:

$(1): \quad y + 2 y' + y = x e^{-x}$

has the general solution:

$y = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$

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:

$p = 2$

$q = 1$

$\map R x = x e^{-x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y + 2 y' + y = 0$

From Linear Second Order ODE: $y + 2 y' + y = 0$, this has the general solution:

$y_g = \paren {C_1 + C_2 x} e^{-x}$

It remains to find a particular solution $y_p$ to $(1)$.

We have that:

$\map R x = x e^{-x}$

and so from the Method of Undetermined Coefficients for Product of Polynomial and Exponential, we assume a solution:

\(\ds y_p\)

\(=\)

\(\ds e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}'\)

\(=\)

\(\ds -e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4} + e^{-x} \paren {3 A_1 x^2 + 2 A_2 x + A_3}\)

\(\ds \)

\(=\)

\(\ds e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4}\)

\(\ds \leadsto \ \ \)

\(\ds {y_p}\)

\(=\)

\(\ds -e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4} + e^{-x} \paren {-3 A_1 x^2 + 2 \paren {3 A_1 - A_2} x + \paren {2 A_2 - A_3} }\)

\(\ds \)

\(=\)

\(\ds e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + 2 A_2 - 2 A_3 + A_4}\)

Substituting in $(1)$:

\(\ds \)

\(\)

\(\ds e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + 2 A_2 - 2 A_3 + A_4}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds 2 e^{-x} \paren {-A_1 x^3 + \paren {3 A_1 - A_2} x^2 + \paren {2 A_2 - A_3} x + A_3 - A_4}\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}\)

\(\ds \)

\(=\)

\(\ds x e^{-x}\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \)

\(\)

\(\ds e^{-x} \paren {A_1 x^3 + \paren {-6 A_1 + A_2} x^2 + \paren {6 A_1 - 4 A_2 + A_3} x + \paren {2 A_2 - 2 A_3 + A_4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds e^{-x} \paren {-2 A_1 x^3 + \paren {6 A_1 - 2 A_2} x^2 + \paren {4 A_2 - 2 A_3} x + \paren {2 A_3 - 2 A_4} }\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds e^{-x} \paren {A_1 x^3 + A_2 x^2 + A_3 x + A_4}\)

\(\ds \)

\(=\)

\(\ds x e^{-x}\)

Hence by equating coefficients:

\(\ds A_1 - 2 A_1 + A_1\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds 0\)

\(\ds \paren {-6 A_1 + A_2} + \paren {6 A_1 - 2 A_2} + A_2\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds 0\)

\(\ds \paren {6 A_1 - 4 A_2 + A_3} + \paren {4 A_2 - 2 A_3} + A_3\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds 6 A_1\)

\(=\)

\(\ds 1\)

\(\ds \paren {2 A_2 - 2 A_3 + A_4} + \paren {2 A_3 - 2 A_4} + A_4\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds A_2\)

\(=\)

\(\ds 0\)

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$y = y_g + y_p = e^{-x} \paren {C_1 + C_2 x + \dfrac {x^3} 6}$

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $9$