Linear Second Order ODE/y'' - 7 y' - 5 y = x^3 - 1
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Theorem
The second order ODE:
- $(1): \quad y - 7 y' - 5 y = x^3 - 1$
has the general solution:
- $y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$
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:
- $p = -7$
- $q = -5$
- $\map R x = x^3 - 1$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $(2): \quad y - 7 y' - 5 y = 0$
From Linear Second Order ODE: $y - 7 y' - 5 y = 0$, this has the general solution:
- $y_g = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$
We have that:
- $\map R x = x^3 - 1$
and it is noted that $x^3 - 1$ is not itself a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Polynomials:
- $y_p = A_0 + A_1 x + A_2 x^2 + A_3 x^3$
for $A_n$ to be determined.
Hence:
\(\ds y_p\) | \(=\) | \(\ds A_0 + A_1 x + A_2 x^2 + A_3 x^3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds A_1 + 2 A_2 x + 3 A_3 x^2\) | Power Rule for Derivatives | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}\) | \(=\) | \(\ds 2 A_2 + 6 A_3 x\) | Power Rule for Derivatives |
Substituting into $(1)$:
\(\ds 2 A_2 + 6 A_3 x - 7 \paren {A_1 + 2 A_2 x + 3 A_3 x^2} - 5 \paren {A_0 + A_1 x + A_2 x^2 + A_3 x^3}\) | \(=\) | \(\ds x^3 - 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds -5 A_3 x^3\) | \(=\) | \(\ds x^3\) | equating powers | ||||||||||
\(\ds \paren {-21 A_3 - 5 A_2} x^2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \paren {6 A_3 - 14 A_2 - 5 A_1} x\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds 2 A_2 - 7 A_1 - 5 A_0\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_3\) | \(=\) | \(\ds -\dfrac 1 5\) | |||||||||||
\(\ds A_2\) | \(=\) | \(\ds -\dfrac {21} 5 A_3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac {21} {25}\) | ||||||||||||
\(\ds 5 A_1\) | \(=\) | \(\ds -\dfrac 6 5 - \dfrac {294} {25}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_1\) | \(=\) | \(\ds -\dfrac {324} {125}\) | |||||||||||
\(\ds 5 A_0\) | \(=\) | \(\ds \dfrac {42} {25} + \dfrac {2268} {125} + 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A_0\) | \(=\) | \(\ds \dfrac {2603} {625}\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.3$ Particular solution: polynomial $\map f x$