Linear Second Order ODE/y'' - 2 y' - 3 y = 64 x exp -x

Theorem

The second order ODE:

$(1): \quad y - 2 y' - 3 y = 64 x e^{-x}$

has the general solution:

$y = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$

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:

$p = -2$

$q = -3$

$\map R x = 64 x e^{-x}$

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

$y - 2 y' - 3 y = 0$

From Linear Second Order ODE: $y - 2 y' - 3 y = 0$, this has the general solution:

$y_g = C_1 e^{3 x} + C_2 e^{-x}$

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:

\(\ds \map {y_1} x\)

\(=\)

\(\ds e^{3 x}\)

\(\ds \map {y_2} x\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds \map { {y_1}'} x\)

\(=\)

\(\ds 3 e^{3 x}\)

Derivative of Exponential Function

\(\ds \map { {y_2}'} x\)

\(=\)

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

Derivative of Exponential Function

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

\(\ds \)

\(=\)

\(\ds e^{3 x} \paren {-e^{-x} } - e^{-x} \paren {3 e^{3 x} }\)

\(\ds \)

\(=\)

\(\ds -e^{2 x} - 3 e^{2 x}\)

\(\ds \)

\(=\)

\(\ds -4 e^{2 x}\)

Hence:

\(\ds v_1\)

\(=\)

\(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)

\(\ds \)

\(=\)

\(\ds \int -\frac {e^{-x} 64 x e^{-x} } {-4 e^{2 x} } \rd x\)

\(\ds \)

\(=\)

\(\ds 16 \int x e^{-4 x} \rd x\)

\(\ds \)

\(=\)

\(\ds 16 \frac {e^{-4 x} } {-4} \paren {x - \frac 1 {-4} }\)

Primitive of $x e^{a x}$

\(\ds \)

\(=\)

\(\ds -4 e^{-4 x} \paren {x + \frac 1 4}\)

\(\ds \)

\(=\)

\(\ds -4 x e^{-4 x} - e^{-4 x}\)

\(\ds v_2\)

\(=\)

\(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)

\(\ds \)

\(=\)

\(\ds \int \frac {e^{3 x} 64 x e^{-x} } {-4 e^{2 x} } \rd x\)

\(\ds \)

\(=\)

\(\ds - 16 \int x \rd x\)

\(\ds \)

\(=\)

\(\ds -8 x^2\)

Primitive of Power

It follows that:

\(\ds y_p\)

\(=\)

\(\ds \paren {-4 x e^{-4 x} - e^{-4 x} } e^{3 x} + \paren {-8 x^2} e^{-x}\)

\(\ds \)

\(=\)

\(\ds -4 x e^{-x} - e^{-x} - 8 x^2 e^{-x}\)

simplifying

\(\ds \)

\(=\)

\(\ds -e^{-x} \paren {8 x^2 + 4 x + 1}\)

simplifying

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

$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$

is the general solution to $(1)$.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $3 \ \text{(c)}$