Linear Second Order ODE/y'' - 4 y' - 5 y = x^2/y(0) = 1, y'(0) = -1

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Theorem

Consider the second order ODE:

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

whose initial conditions are:

$y = 1$ when $x = 0$
$y' = -1$ when $x = 0$


$(1)$ has the particular solution:

$y = \dfrac {e^{5 x} } {375} + \dfrac {4 e^{-x} } 3 - \dfrac {x^2} 5 + \dfrac {8 x} {25} - \dfrac {42} {125}$


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:

$p = -4$
$q = -5$
$\map R x = x^2$


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

$(2): \quad y - 4 y' - 5 y = 0$

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

$y_g = C_1 e^{5 x} + C_2 e^{-x}$


We have that:

$\map R x = x^2$

and it is noted that $x^2$ is not itself a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for Polynomials:

$y_p = A_0 + A_1 x + A_2 x^2$

for $A_n$ to be determined.


Hence:

\(\ds y_p\) \(=\) \(\ds A_0 + A_1 x + A_2 x^2\)
\(\ds \leadsto \ \ \) \(\ds {y_p}'\) \(=\) \(\ds A_1 + 2 A_2 x\) Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds {y_p}\) \(=\) \(\ds 2 A_2\) Power Rule for Derivatives


Substituting into $(1)$:

\(\ds 2 A_2 - 4 \paren {A_1 + 2 A_2 x} - 5 \paren {A_0 + A_1 x + A_2 x^2}\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds -5 A_2 x^2\) \(=\) \(\ds x^2\) equating coefficients of powers
\(\ds \paren {-8 A_2 - 5 A_1} x\) \(=\) \(\ds 0\)
\(\ds 2 A_2 - 4 A_1 - 5 A_0\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A_2\) \(=\) \(\ds -\dfrac 1 5\)
\(\ds 5 A_1\) \(=\) \(\ds \dfrac 8 5\)
\(\ds \leadsto \ \ \) \(\ds A_1\) \(=\) \(\ds \dfrac 8 {25}\)
\(\ds 5 A_0\) \(=\) \(\ds -\dfrac 2 5 - \dfrac {4 \times 8} {25}\)
\(\ds \leadsto \ \ \) \(\ds A_0\) \(=\) \(\ds -\dfrac {42} {125}\)

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

$y = y_g + y_p = C_1 e^{5 x} + C_2 e^{-x} - \dfrac 1 5 x^2 + \dfrac 8 {25} x - \dfrac {42} {125}$


Differentiating with respect to $x$:

$y' = 5 C_1 e^{5 x} - C_2 e^{-x} - \dfrac 2 5 x + \dfrac 8 {25}$

Substituting the initial conditions, remembering that $e^0 = 1$:

\(\ds 1\) \(=\) \(\ds C_1 + C_2 - \dfrac {42} {125}\) setting $y = 1$ when $x = 0$
\(\ds -1\) \(=\) \(\ds 5 C_1 - C_2 + \dfrac 8 {25}\) setting $y' = -1$ when $x = 0$
\(\ds \leadsto \ \ \) \(\ds -\dfrac 2 {125} + 6 C_1\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds C_1\) \(=\) \(\ds \dfrac 1 {375}\)
\(\ds \leadsto \ \ \) \(\ds C_2\) \(=\) \(\ds 1 - 5 A + \dfrac 8 {25}\)
\(\ds \) \(=\) \(\ds 1 + \dfrac 1 {75} + \dfrac 8 {27}\)
\(\ds \) \(=\) \(\ds \dfrac 4 3\)

The result follows.

$\blacksquare$


Sources

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