Internal problem ID [1566]
Internal file name [OUTPUT/1567_Sunday_June_05_2022_02_22_44_AM_43420082/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 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 69.
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, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }-2 y^{\prime \prime }-5 y^{\prime }+6 y=2 \,{\mathrm e}^{x} \left (1-6 x \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 7, y^{\prime \prime }\left (0\right ) = 9] \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 \[ y^{\prime \prime \prime }-2 y^{\prime \prime }-5 y^{\prime }+6 y = 0 \] The characteristic equation is \[ \lambda ^{3}-2 \lambda ^{2}-5 \lambda +6 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 3\\ \lambda _3 &= -2 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-2 x} \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{3 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-2 y^{\prime \prime }-5 y^{\prime }+6 y = 2 \,{\mathrm e}^{x} \left (1-6 x \right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 \,{\mathrm e}^{x} \left (1-6 x \right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{-2 x}, {\mathrm e}^{3 x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] 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. \[ y_p = A_{1} x \,{\mathrm e}^{x}+A_{2} x^{2} {\mathrm e}^{x} \] 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 \[ -6 A_{1} {\mathrm e}^{x}+2 A_{2} {\mathrm e}^{x}-12 A_{2} x \,{\mathrm e}^{x} = 2 \,{\mathrm e}^{x} \left (1-6 x \right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = x^{2} {\mathrm e}^{x} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3}\right ) + \left (x^{2} {\mathrm e}^{x}\right ) \\ \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}+c_{2} {\mathrm e}^{x}+{\mathrm e}^{3 x} c_{3} +x^{2} {\mathrm e}^{x} \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}+c_{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{3 x} c_{3} +2 x \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} \end {align*}
substituting \(y^{\prime } = 7\) and \(x = 0\) in the above gives \begin {align*} 7 = -2 c_{1} +c_{2} +3 c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 4 c_{1} {\mathrm e}^{-2 x}+c_{2} {\mathrm e}^{x}+9 \,{\mathrm e}^{3 x} c_{3} +4 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{x} \end {align*}
substituting \(y^{\prime \prime } = 9\) and \(x = 0\) in the above gives \begin {align*} 9 = 4 c_{1} +c_{2} +9 c_{3} +2\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}&=-1\\ c_{2}&=2\\ c_{3}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x}-{\mathrm e}^{-2 x}+{\mathrm e}^{3 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x}-{\mathrm e}^{-2 x}+{\mathrm e}^{3 x} \\ \end{align*}
Verification of solutions
\[ y = x^{2} {\mathrm e}^{x}+2 \,{\mathrm e}^{x}-{\mathrm e}^{-2 x}+{\mathrm e}^{3 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-2 y^{\prime \prime }-5 y^{\prime }+6 y=2 \,{\mathrm e}^{x} \left (1-6 x \right ), y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=9\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-6 y-12 x \,{\mathrm e}^{x}+2 y^{\prime \prime }+5 y^{\prime }+2 \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-2 y^{\prime \prime }-5 y^{\prime }+6 y=-2 \,{\mathrm e}^{x} \left (-1+6 x \right ) \\ \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 )=-12 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}+2 y_{3}\left (x \right )+5 y_{2}\left (x \right )-6 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 )=-12 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x}+2 y_{3}\left (x \right )+5 y_{2}\left (x \right )-6 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 \\ -6 & 5 & 2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ -12 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -12 x \,{\mathrm e}^{x}+2 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & 5 & 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 )+{\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 [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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 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}}_{2}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 x}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 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 )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\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} \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{9} \\ -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{x} & {\mathrm e}^{3 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 \frac {1}{\phi \left (0\right )} \\ {} & \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} \frac {{\mathrm e}^{-2 x}}{4} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{9} \\ -\frac {{\mathrm e}^{-2 x}}{2} & {\mathrm e}^{x} & \frac {{\mathrm e}^{3 x}}{3} \\ {\mathrm e}^{-2 x} & {\mathrm e}^{x} & {\mathrm e}^{3 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} \frac {1}{4} & 1 & \frac {1}{9} \\ -\frac {1}{2} & 1 & \frac {1}{3} \\ 1 & 1 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} -\frac {\left ({\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (3 \,{\mathrm e}^{5 x}+5 \,{\mathrm e}^{3 x}-8\right ) {\mathrm e}^{-2 x}}{30} & \frac {\left (3 \,{\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{30} \\ -\frac {\left (3 \,{\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (9 \,{\mathrm e}^{5 x}+5 \,{\mathrm e}^{3 x}+16\right ) {\mathrm e}^{-2 x}}{30} & \frac {\left (9 \,{\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}-4\right ) {\mathrm e}^{-2 x}}{30} \\ -\frac {\left (9 \,{\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}-4\right ) {\mathrm e}^{-2 x}}{5} & \frac {\left (27 \,{\mathrm e}^{5 x}+5 \,{\mathrm e}^{3 x}-32\right ) {\mathrm e}^{-2 x}}{30} & \frac {\left (27 \,{\mathrm e}^{5 x}-5 \,{\mathrm e}^{3 x}+8\right ) {\mathrm e}^{-2 x}}{30} \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} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \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 {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\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}\frac {1}{\Phi \left (s \right )}\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}\frac {1}{\Phi \left (s \right )}\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 {\left (3 \,{\mathrm e}^{5 x}-15 x^{2} {\mathrm e}^{3 x}-5 \,{\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (15 x^{2}+30 x +5\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{15}+\frac {\left (-9 \,{\mathrm e}^{5 x}+4\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (15 x^{2}+60 x +35\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{15}+\frac {\left (-27 \,{\mathrm e}^{5 x}-8\right ) {\mathrm e}^{-2 x}}{15} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} -\frac {\left (3 \,{\mathrm e}^{5 x}-15 x^{2} {\mathrm e}^{3 x}-5 \,{\mathrm e}^{3 x}+2\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (15 x^{2}+30 x +5\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{15}+\frac {\left (-9 \,{\mathrm e}^{5 x}+4\right ) {\mathrm e}^{-2 x}}{15} \\ \frac {\left (15 x^{2}+60 x +35\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{15}+\frac {\left (-27 \,{\mathrm e}^{5 x}-8\right ) {\mathrm e}^{-2 x}}{15} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (180 x^{2}+180 c_{2} +60\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{180}+\frac {\left (20 \,{\mathrm e}^{5 x} c_{3} +45 c_{1} -36 \,{\mathrm e}^{5 x}-24\right ) {\mathrm e}^{-2 x}}{180} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{2} +\frac {c_{3}}{9}+\frac {c_{1}}{4} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 x \,{\mathrm e}^{-2 x} {\mathrm e}^{3 x}+\frac {\left (180 x^{2}+180 c_{2} +60\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{180}+\frac {\left (100 \,{\mathrm e}^{5 x} c_{3} -180 \,{\mathrm e}^{5 x}\right ) {\mathrm e}^{-2 x}}{180}-\frac {\left (20 \,{\mathrm e}^{5 x} c_{3} +45 c_{1} -36 \,{\mathrm e}^{5 x}-24\right ) {\mathrm e}^{-2 x}}{90} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=7 \\ {} & {} & 7=c_{2} +\frac {c_{3}}{3}-\frac {c_{1}}{2} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=2 \,{\mathrm e}^{-2 x} {\mathrm e}^{3 x}+4 x \,{\mathrm e}^{-2 x} {\mathrm e}^{3 x}+\frac {\left (180 x^{2}+180 c_{2} +60\right ) {\mathrm e}^{-2 x} {\mathrm e}^{3 x}}{180}+\frac {\left (500 \,{\mathrm e}^{5 x} c_{3} -900 \,{\mathrm e}^{5 x}\right ) {\mathrm e}^{-2 x}}{180}-\frac {\left (100 \,{\mathrm e}^{5 x} c_{3} -180 \,{\mathrm e}^{5 x}\right ) {\mathrm e}^{-2 x}}{45}+\frac {\left (20 \,{\mathrm e}^{5 x} c_{3} +45 c_{1} -36 \,{\mathrm e}^{5 x}-24\right ) {\mathrm e}^{-2 x}}{45} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=9 \\ {} & {} & 9=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-\frac {52}{15}, c_{2} =\frac {5}{3}, c_{3} =\frac {54}{5}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left ({\mathrm e}^{5 x}+x^{2} {\mathrm e}^{3 x}+2 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x} \end {array} \]
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`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 29
dsolve([diff(y(x),x$3)-2*diff(y(x),x$2)-5*diff(y(x),x)+6*y(x)=2*exp(x)*(1-6*x),y(0) = 2, D(y)(0) = 7, (D@@2)(y)(0) = 9],y(x), singsol=all)
\[ y \left (x \right ) = \left ({\mathrm e}^{5 x}+x^{2} {\mathrm e}^{3 x}+2 \,{\mathrm e}^{3 x}-1\right ) {\mathrm e}^{-2 x} \]
✓ Solution by Mathematica
Time used: 0.061 (sec). Leaf size: 27
DSolve[{y'''[x]-2*y''[x]-5*y'[x]+6*y[x]==2*Exp[x]*(1-6*x),{y[0]==2,y'[0]==7,y''[0]==9}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x \left (x^2+2\right )-e^{-2 x}+e^{3 x} \]