Linear Second Order ODE/y'' + 4 y' + 5 y = 2 exp -2 x

Theorem

The second order ODE:

$(1): \quad y + 4 y' + 5 y = 2 e^{-2 x}$

has the general solution:

$y = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x + 2}$

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:

$p = 4$

$q = 5$

$\map R x = 2 e^{-2 x}$

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

$y + 4 y' + 5 y = 0$

From Linear Second Order ODE: $y + 4 y' + 5 y = 0$, this has the general solution:

$y_g = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$

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

We have that:

$\map R x = 2 e^{-2 x}$

and so from the Method of Undetermined Coefficients for the Exponential Function:

$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$

where:

$K = 2$

$a = -2$

$p = 4$

$q = 5$

Hence:

\(\ds y_p\)

\(=\)

\(\ds \dfrac {2 e^{-2 x} } {\paren {-2}^2 + 4 \times \paren {-2} + 5}\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 e^{-2 x} } {4 - 8 + 5}\)

\(\ds \)

\(=\)

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

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

$y = y_g + y_p = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x + 2}$

is the general solution to $(1)$.

$\blacksquare$

Sources

• 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: Problems for Chapter $1$: $5$