Internal problem ID [1479]
Internal file name [OUTPUT/1480_Sunday_June_05_2022_02_19_14_AM_82956636/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.2. constant
coefficient. Page 483
Problem number: section 9.2, problem 15.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_3rd_order, _missing_x]]
\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }-8 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = -2, y^{\prime \prime }\left (0\right ) = 2] \end {align*}
The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}+4 \lambda -8 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 2\\ \lambda _2 &= 2 i\\ \lambda _3 &= -2 i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{2} +{\mathrm e}^{2 i x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{2 x}\\ y_2 &= {\mathrm e}^{-2 i x}\\ y_3 &= {\mathrm e}^{2 i x} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{2 x}+{\mathrm e}^{-2 i x} c_{2} +{\mathrm e}^{2 i x} c_{3} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{1} {\mathrm e}^{2 x}-2 i {\mathrm e}^{-2 i x} c_{2} +2 i {\mathrm e}^{2 i x} c_{3} \end {align*}
substituting \(y^{\prime } = -2\) and \(x = 0\) in the above gives \begin {align*} -2 = -2 c_{2} i+2 c_{3} i+2 c_{1}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 4 c_{1} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{-2 i x} c_{2} -4 \,{\mathrm e}^{2 i x} c_{3} \end {align*}
substituting \(y^{\prime \prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = 4 c_{1} -4 c_{2} -4 c_{3}\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&={\frac {5}{4}}\\ c_{2}&=\frac {3}{8}-\frac {9 i}{8}\\ c_{3}&=\frac {3}{8}+\frac {9 i}{8} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {5 \,{\mathrm e}^{2 x}}{4}+\frac {3 \cos \left (2 x \right )}{4}-\frac {9 \sin \left (2 x \right )}{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {5 \,{\mathrm e}^{2 x}}{4}+\frac {3 \cos \left (2 x \right )}{4}-\frac {9 \sin \left (2 x \right )}{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {5 \,{\mathrm e}^{2 x}}{4}+\frac {3 \cos \left (2 x \right )}{4}-\frac {9 \sin \left (2 x \right )}{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-2 y^{\prime \prime }+4 y^{\prime }-8 y=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-2, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \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 \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=2 y_{3}\left (x \right )-4 y_{2}\left (x \right )+8 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=2 y_{3}\left (x \right )-4 y_{2}\left (x \right )+8 y_{1}\left (x \right )\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} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 8 & -4 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 8 & -4 & 2 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} x}\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right )\cdot \left [\begin {array}{c} -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4}+\frac {\mathrm {I} \sin \left (2 x \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right )\right ) \\ \cos \left (2 x \right )-\mathrm {I} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\frac {\cos \left (2 x \right )}{4} \\ \frac {\sin \left (2 x \right )}{2} \\ \cos \left (2 x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \frac {\sin \left (2 x \right )}{4} \\ \frac {\cos \left (2 x \right )}{2} \\ -\sin \left (2 x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{2} \cos \left (2 x \right )}{4}+\frac {c_{3} \sin \left (2 x \right )}{4} \\ \frac {c_{2} \sin \left (2 x \right )}{2}+\frac {c_{3} \cos \left (2 x \right )}{2} \\ c_{2} \cos \left (2 x \right )-c_{3} \sin \left (2 x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {c_{1} {\mathrm e}^{2 x}}{4}+\frac {c_{3} \sin \left (2 x \right )}{4}-\frac {c_{2} \cos \left (2 x \right )}{4} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=\frac {c_{1}}{4}-\frac {c_{2}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1} {\mathrm e}^{2 x}}{2}+\frac {c_{3} \cos \left (2 x \right )}{2}+\frac {c_{2} \sin \left (2 x \right )}{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=-2 \\ {} & {} & -2=\frac {c_{1}}{2}+\frac {c_{3}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=c_{1} {\mathrm e}^{2 x}-c_{3} \sin \left (2 x \right )+c_{2} \cos \left (2 x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=c_{1} +c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =5, c_{2} =-3, c_{3} =-9\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5 \,{\mathrm e}^{2 x}}{4}+\frac {3 \cos \left (2 x \right )}{4}-\frac {9 \sin \left (2 x \right )}{4} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 23
dsolve([diff(y(x),x$3)-2*diff(y(x),x$2)+4*diff(y(x),x)-8*y(x)=0,y(0) = 2, D(y)(0) = -2, (D@@2)(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = \frac {5 \,{\mathrm e}^{2 x}}{4}-\frac {9 \sin \left (2 x \right )}{4}+\frac {3 \cos \left (2 x \right )}{4} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 21
DSolve[{y'''[x]-2*y''[x]+4*y'[x]-8*y[x]==0,{y[0]==2,y'[0]==-2,y''[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{2 x}-2 \sin (2 x)+\cos (2 x) \]