19.71 problem section 9.3, problem 71
Internal
problem
ID
[2218]
Book
:
Elementary
differential
equations
with
boundary
value
problems.
William
F.
Trench.
Brooks/Cole
2001
Section
:
Chapter
9
Introduction
to
Linear
Higher
Order
Equations.
Section
9.3.
Undetermined
Coefficients
for
Higher
Order
Equations.
Page
495
Problem
number
:
section
9.3,
problem
71
Date
solved
:
Thursday, October 17, 2024 at 02:21:49 AM
CAS
classification
:
[[_3rd_order, _linear, _nonhomogeneous]]
Solve
\begin{align*} 4 y^{\prime \prime \prime }-3 y^{\prime }-y&={\mathrm e}^{-\frac {x}{2}} \left (2-3 x \right ) \end{align*}
With initial conditions
\begin{align*} y \left (0\right )&=-1\\ y^{\prime }\left (0\right )&=15\\ y^{\prime \prime }\left (0\right )&=-17 \end{align*}
19.71.1 Solved as higher order constant coeff ode
Time used: 0.112 (sec)
The characteristic equation is
\[ 4 \lambda ^{3}-3 \lambda -1 = 0 \]
The roots of the above equation are
\begin{align*} \lambda _1 &= 1\\ \lambda _2 &= -{\frac {1}{2}}\\ \lambda _3 &= -{\frac {1}{2}} \end{align*}
Therefore the homogeneous solution is
\[ y_h(x)={\mathrm e}^{x} c_1 +{\mathrm e}^{-\frac {x}{2}} c_2 +x \,{\mathrm e}^{-\frac {x}{2}} c_3 \]
The fundamental set of solutions for the
homogeneous solution are the following
\begin{align*} y_1 &= {\mathrm e}^{x}\\ y_2 &= {\mathrm e}^{-\frac {x}{2}}\\ y_3 &= {\mathrm e}^{-\frac {x}{2}} x \end{align*}
This is higher order nonhomogeneous ODE. Let the solution be
\[ y = y_h + y_p \]
Where \(y_h\) is the solution to
the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the
solution to
\[ 4 y^{\prime \prime \prime }-3 y^{\prime }-y = 0 \]
Now the particular solution to the given ODE is found
\[
4 y^{\prime \prime \prime }-3 y^{\prime }-y = -{\mathrm e}^{-\frac {x}{2}} \left (-2+3 x \right )
\]
The particular solution
is now found using the method of undetermined coefficients.
Looking at the RHS of the ode, which is
\[ -{\mathrm e}^{-\frac {x}{2}} \left (-2+3 x \right ) \]
Shows that the corresponding undetermined set of
the basis functions (UC_set) for the trial solution is
\[ \left [\left \{{\mathrm e}^{-\frac {x}{2}} x, {\mathrm e}^{-\frac {x}{2}}\right \}\right ] \]
While the set of the basis functions for
the homogeneous solution found earlier is
\[ \left \{{\mathrm e}^{-\frac {x}{2}} x, {\mathrm e}^{x}, {\mathrm e}^{-\frac {x}{2}}\right \} \]
Since \({\mathrm e}^{-\frac {x}{2}}\) is duplicated in the UC_set, then this basis
is multiplied by extra \(x\) . The UC_set becomes
\[ \left [\left \{x^{2} {\mathrm e}^{-\frac {x}{2}}, {\mathrm e}^{-\frac {x}{2}} x\right \}\right ] \]
Since \({\mathrm e}^{-\frac {x}{2}} x\) is duplicated in the UC_set,
then this basis is multiplied by extra \(x\) . The UC_set becomes
\[ \left [\left \{x^{2} {\mathrm e}^{-\frac {x}{2}}, x^{3} {\mathrm e}^{-\frac {x}{2}}\right \}\right ] \]
Since there was
duplication between the basis functions in the UC_set and the basis functions of the
homogeneous solution, the trial solution is a linear combination of all the basis
function in the above updated UC_set.
\[
y_p = A_{1} x^{2} {\mathrm e}^{-\frac {x}{2}}+A_{2} x^{3} {\mathrm e}^{-\frac {x}{2}}
\]
The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the
above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the
trial solution into the ODE and simplifying gives
\[
-12 A_{1} {\mathrm e}^{-\frac {x}{2}}+24 A_{2} {\mathrm e}^{-\frac {x}{2}}-36 A_{2} x \,{\mathrm e}^{-\frac {x}{2}} = {\mathrm e}^{-\frac {x}{2}} \left (2-3 x \right )
\]
Solving for the unknowns by
comparing coefficients results in
\[ \left [A_{1} = 0, A_{2} = {\frac {1}{12}}\right ] \]
Substituting the above back in the above trial
solution \(y_p\) , gives the particular solution
\[
y_p = \frac {x^{3} {\mathrm e}^{-\frac {x}{2}}}{12}
\]
Therefore the general solution is
\begin{align*}
y &= y_h + y_p \\
&= \left ({\mathrm e}^{x} c_1 +{\mathrm e}^{-\frac {x}{2}} c_2 +x \,{\mathrm e}^{-\frac {x}{2}} c_3\right ) + \left (\frac {x^{3} {\mathrm e}^{-\frac {x}{2}}}{12}\right ) \\
\end{align*}
Solving
for constants of integration using given initial conditions, the solution becomes
\begin{align*}
y &= -{\mathrm e}^{x}+16 \,{\mathrm e}^{-\frac {x}{2}} x +\frac {x^{3} {\mathrm e}^{-\frac {x}{2}}}{12} \\
\end{align*}
Figure 692: Solution plot
\(y = -{\mathrm e}^{x}+16 \,{\mathrm e}^{-\frac {x}{2}} x +\frac {x^{3} {\mathrm e}^{-\frac {x}{2}}}{12}\)
19.71.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 \frac {d^{3}}{d x^{3}}y \left (x \right )-3 \frac {d}{d x}y \left (x \right )-y \left (x \right )={\mathrm e}^{-\frac {x}{2}} \left (-3 x +2\right ), y \left (0\right )=-1, \left (\frac {d}{d x}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=15, \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-17\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )=\frac {y \left (x \right )}{4}+\frac {3 \frac {d}{d x}y \left (x \right )}{4}-\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}+\frac {{\mathrm e}^{-\frac {x}{2}}}{2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )-\frac {3 \frac {d}{d x}y \left (x \right )}{4}-\frac {y \left (x \right )}{4}=-\frac {{\mathrm e}^{-\frac {x}{2}} \left (3 x -2\right )}{4} \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=\frac {d}{d x}y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}y_{3}\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y_{3}\left (x \right )=-\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}+\frac {{\mathrm e}^{-\frac {x}{2}}}{2}+\frac {3 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{4} \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=\frac {d}{d x}y_{1}\left (x \right ), y_{3}\left (x \right )=\frac {d}{d x}y_{2}\left (x \right ), \frac {d}{d x}y_{3}\left (x \right )=-\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}+\frac {{\mathrm e}^{-\frac {x}{2}}}{2}+\frac {3 y_{2}\left (x \right )}{4}+\frac {y_{1}\left (x \right )}{4}\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{4} & \frac {3}{4} & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}+\frac {{\mathrm e}^{-\frac {x}{2}}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}+\frac {{\mathrm e}^{-\frac {x}{2}}}{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{4} & \frac {3}{4} & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ], \left [-\frac {1}{2}, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-\frac {1}{2}, \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =-\frac {1}{2}\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =-\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac {1}{4} & \frac {3}{4} & 0 \end {array}\right ]--\frac {1}{2}\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} 8 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -\frac {1}{2} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-\frac {x}{2}}\cdot \left (x \cdot \left [\begin {array}{c} 4 \\ -2 \\ 1 \end {array}\right ]+\left [\begin {array}{c} 8 \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (8+4 x \right ) & {\mathrm e}^{x} \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 \,{\mathrm e}^{-\frac {x}{2}} x & {\mathrm e}^{x} \\ {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} x & {\mathrm e}^{x} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \phi \left (0\right )^{-1} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} 4 \,{\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \left (8+4 x \right ) & {\mathrm e}^{x} \\ -2 \,{\mathrm e}^{-\frac {x}{2}} & -2 \,{\mathrm e}^{-\frac {x}{2}} x & {\mathrm e}^{x} \\ {\mathrm e}^{-\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} x & {\mathrm e}^{x} \end {array}\right ]\cdot \left [\begin {array}{ccc} 4 & 8 & 1 \\ -2 & 0 & 1 \\ 1 & 0 & 1 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {{\mathrm e}^{-\frac {x}{2}} \left (x +2\right )}{2} & \frac {{\mathrm e}^{-\frac {x}{2}} \left (3 x -2\right )}{6}+\frac {{\mathrm e}^{x}}{3} & -\frac {2 \,{\mathrm e}^{-\frac {x}{2}}}{3}-{\mathrm e}^{-\frac {x}{2}} x +\frac {2 \,{\mathrm e}^{x}}{3} \\ -\frac {{\mathrm e}^{-\frac {x}{2}} x}{4} & \frac {\left (-3 x +8\right ) {\mathrm e}^{-\frac {x}{2}}}{12}+\frac {{\mathrm e}^{x}}{3} & \frac {\left (3 x -4\right ) {\mathrm e}^{-\frac {x}{2}}}{6}+\frac {2 \,{\mathrm e}^{x}}{3} \\ \frac {{\mathrm e}^{-\frac {x}{2}} x}{8} & \frac {\left (3 x -8\right ) {\mathrm e}^{-\frac {x}{2}}}{24}+\frac {{\mathrm e}^{x}}{3} & \frac {\left (4-3 x \right ) {\mathrm e}^{-\frac {x}{2}}}{12}+\frac {2 \,{\mathrm e}^{x}}{3} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )=\Phi \left (x \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {{\mathrm e}^{-\frac {x}{2}} x^{3}}{8} \\ -\frac {{\mathrm e}^{-\frac {x}{2}} x^{2} \left (-6+x \right )}{16} \\ \frac {{\mathrm e}^{-\frac {x}{2}} x \left (x^{2}-6 x +16\right )}{32} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}\left (x \right )+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} \frac {{\mathrm e}^{-\frac {x}{2}} x^{3}}{8} \\ -\frac {{\mathrm e}^{-\frac {x}{2}} x^{2} \left (-6+x \right )}{16} \\ \frac {{\mathrm e}^{-\frac {x}{2}} x \left (x^{2}-6 x +16\right )}{32} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (32 \mathit {C2} \left (x +2\right )+x^{3}+32 \mathit {C1} \right ) {\mathrm e}^{-\frac {x}{2}}}{8}+\mathit {C3} \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=8 \mathit {C2} +4 \mathit {C1} +\mathit {C3} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\left (3 x^{2}+32 \mathit {C2} \right ) {\mathrm e}^{-\frac {x}{2}}}{8}-\frac {\left (32 \mathit {C2} \left (x +2\right )+x^{3}+32 \mathit {C1} \right ) {\mathrm e}^{-\frac {x}{2}}}{16}+\mathit {C3} \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d x}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=15 \\ {} & {} & 15=-2 \mathit {C1} +\mathit {C3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {3 \,{\mathrm e}^{-\frac {x}{2}} x}{4}-\frac {\left (3 x^{2}+32 \mathit {C2} \right ) {\mathrm e}^{-\frac {x}{2}}}{8}+\frac {\left (32 \mathit {C2} \left (x +2\right )+x^{3}+32 \mathit {C1} \right ) {\mathrm e}^{-\frac {x}{2}}}{32}+\mathit {C3} \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-17 \\ {} & {} & -17=-2 \mathit {C2} +\mathit {C1} +\mathit {C3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{\mathit {C1} =-8, \mathit {C2} =4, \mathit {C3} =-1\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (x^{3}+128 x \right ) {\mathrm e}^{-\frac {x}{2}}}{8}-{\mathrm e}^{x} \end {array} \]
19.71.3 Maple trace
` Methods for third order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 3; linear nonhomogeneous with symmetry [0,1]
trying high order linear exact nonhomogeneous
trying differential order: 3; missing the dependent variable
checking if the LODE has constant coefficients
<- constant coefficients successful `
19.71.4 Maple dsolve solution
Solving time : 0.023
(sec)
Leaf size : 22
dsolve ([4* diff ( diff ( diff ( y ( x ), x ), x ), x )-3* diff ( y ( x ), x )- y ( x ) = exp (-1/2* x )*(2-3* x ),
op ([ y (0) = -1, D ( y )(0) = 15, (D@@2)(y)(0) = -17])],y(x),singsol=all)
\[
y = \frac {\left (x^{3}+192 x \right ) {\mathrm e}^{-\frac {x}{2}}}{12}-{\mathrm e}^{x}
\]
19.71.5 Mathematica DSolve solution
Solving time : 0.03
(sec)
Leaf size : 35
DSolve [{4* D [ y [ x ],{ x ,3}]-0* D [ y [ x ],{ x ,2}]-3* D [ y [ x ], x ]-1* y [ x ]== Exp [- x /2]*(2-3* x ),{ y [0]==2, Derivative [1][ y ][0] ==0, Derivative [2][y][0] ==0}},
y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to \frac {1}{36} e^{-x/2} \left (3 x^3+24 x+8 e^{3 x/2}+64\right )
\]