Internal problem ID [11819]
Internal file name [OUTPUT/11829_Thursday_April_11_2024_08_51_25_PM_72330630/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 4, Section 4.3. The method of undetermined coefficients. Exercises page
151
Problem number: 46.
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_y]]
\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y^{\prime }=x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2}} \] 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 \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-3 \lambda ^{2}+2 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 2\\ \lambda _3 &= 1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= {\mathrm e}^{x} \\ y_3 &= {\mathrm e}^{2 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y^{\prime } = x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}, \{1, x, x^{2}\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{x}, {\mathrm e}^{2 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}, \{x, x^{2}, x^{3}\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{2 x} x, {\mathrm e}^{2 x}\}, \{x, x^{2}, x^{3}\}, \{x \,{\mathrm e}^{x}, x^{3} {\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] Since \({\mathrm e}^{2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{2 x}, {\mathrm e}^{2 x} x\}, \{x, x^{2}, x^{3}\}, \{x \,{\mathrm e}^{x}, x^{3} {\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \] 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. \[ y_p = A_{1} x^{2} {\mathrm e}^{2 x}+A_{2} {\mathrm e}^{2 x} x +A_{3} x +A_{4} x^{2}+A_{5} x^{3}+A_{6} x \,{\mathrm e}^{x}+A_{7} x^{3} {\mathrm e}^{x}+A_{8} x^{2} {\mathrm e}^{x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}\}\) 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 \[ 2 A_{3}-6 A_{4}+6 A_{5}-3 A_{7} x^{2} {\mathrm e}^{x}-2 A_{8} x \,{\mathrm e}^{x}+4 A_{1} x \,{\mathrm e}^{2 x}-18 A_{5} x +4 A_{4} x +6 A_{5} x^{2}+6 A_{1} {\mathrm e}^{2 x}+2 A_{2} {\mathrm e}^{2 x}-A_{6} {\mathrm e}^{x}+6 A_{7} {\mathrm e}^{x} = x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {3}{4}}, A_{2} = -{\frac {9}{4}}, A_{3} = {\frac {35}{4}}, A_{4} = {\frac {15}{4}}, A_{5} = {\frac {5}{6}}, A_{6} = -2, A_{7} = -{\frac {1}{3}}, A_{8} = 0\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {3 x^{2} {\mathrm e}^{2 x}}{4}-\frac {9 \,{\mathrm e}^{2 x} x}{4}+\frac {35 x}{4}+\frac {15 x^{2}}{4}+\frac {5 x^{3}}{6}-2 x \,{\mathrm e}^{x}-\frac {x^{3} {\mathrm e}^{x}}{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3}\right ) + \left (\frac {3 x^{2} {\mathrm e}^{2 x}}{4}-\frac {9 \,{\mathrm e}^{2 x} x}{4}+\frac {35 x}{4}+\frac {15 x^{2}}{4}+\frac {5 x^{3}}{6}-2 x \,{\mathrm e}^{x}-\frac {x^{3} {\mathrm e}^{x}}{3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} +\frac {3 x^{2} {\mathrm e}^{2 x}}{4}-\frac {9 \,{\mathrm e}^{2 x} x}{4}+\frac {35 x}{4}+\frac {15 x^{2}}{4}+\frac {5 x^{3}}{6}-2 x \,{\mathrm e}^{x}-\frac {x^{3} {\mathrm e}^{x}}{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} +c_{2} {\mathrm e}^{x}+{\mathrm e}^{2 x} c_{3} +\frac {3 x^{2} {\mathrm e}^{2 x}}{4}-\frac {9 \,{\mathrm e}^{2 x} x}{4}+\frac {35 x}{4}+\frac {15 x^{2}}{4}+\frac {5 x^{3}}{6}-2 x \,{\mathrm e}^{x}-\frac {x^{3} {\mathrm e}^{x}}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }-3 y^{\prime \prime }+2 y^{\prime }=x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \\ \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 )=x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2}+3 y_{3}\left (x \right )-2 y_{2}\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 )=x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2}+3 y_{3}\left (x \right )-2 y_{2}\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 \\ 0 & -2 & 3 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ x^{2} {\mathrm e}^{x}+3 \,{\mathrm e}^{2 x} x +5 x^{2} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & 3 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \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 [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}}_{3}={\mathrm e}^{2 x}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 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} 1 & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ 0 & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ 0 & {\mathrm e}^{x} & {\mathrm e}^{2 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} 1 & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{4} \\ 0 & {\mathrm e}^{x} & \frac {{\mathrm e}^{2 x}}{2} \\ 0 & {\mathrm e}^{x} & {\mathrm e}^{2 x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & \frac {1}{4} \\ 0 & 1 & \frac {1}{2} \\ 0 & 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} 1 & -\frac {3}{2}+2 \,{\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2} & \frac {1}{2}-{\mathrm e}^{x}+\frac {{\mathrm e}^{2 x}}{2} \\ 0 & 2 \,{\mathrm e}^{x}-{\mathrm e}^{2 x} & -{\mathrm e}^{x}+{\mathrm e}^{2 x} \\ 0 & 2 \,{\mathrm e}^{x}-2 \,{\mathrm e}^{2 x} & -{\mathrm e}^{x}+2 \,{\mathrm e}^{2 x} \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 {35}{4}+\frac {\left (3 x^{2}-9 x +17\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (-x^{3}-6 x -39\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{3}}{6}+\frac {15 x^{2}}{4}+\frac {35 x}{4} \\ \frac {35}{4}+\frac {\left (6 x^{2}-12 x +25\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (-x^{3}-3 x^{2}-6 x -45\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{2}}{2}+\frac {15 x}{2} \\ \frac {15}{2}+\frac {\left (6 x^{2}-6 x +19\right ) {\mathrm e}^{2 x}}{2}+\frac {\left (-x^{3}-6 x^{2}-12 x -51\right ) {\mathrm e}^{x}}{3}+5 x \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 {35}{4}+\frac {\left (3 x^{2}-9 x +17\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (-x^{3}-6 x -39\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{3}}{6}+\frac {15 x^{2}}{4}+\frac {35 x}{4} \\ \frac {35}{4}+\frac {\left (6 x^{2}-12 x +25\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (-x^{3}-3 x^{2}-6 x -45\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{2}}{2}+\frac {15 x}{2} \\ \frac {15}{2}+\frac {\left (6 x^{2}-6 x +19\right ) {\mathrm e}^{2 x}}{2}+\frac {\left (-x^{3}-6 x^{2}-12 x -51\right ) {\mathrm e}^{x}}{3}+5 x \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {35}{4}+\frac {\left (3 x^{2}+c_{3} -9 x +17\right ) {\mathrm e}^{2 x}}{4}+\frac {\left (-x^{3}+3 c_{2} -6 x -39\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{3}}{6}+\frac {15 x^{2}}{4}+\frac {35 x}{4}+c_{1} \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] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = _a^2*exp(_a)+3*exp(2*_a)*_a+5*_a^2+3*(diff(_b(_a), _a))-2*_b(_a), _b( Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] <- double symmetry of the form [xi=0, eta=F(x)] successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 55
dsolve(diff(y(x),x$3)-3*diff(y(x),x$2)+2*diff(y(x),x)=x^2*exp(x)+3*x*exp(2*x)+5*x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (6 x^{2}+4 c_{1} -18 x +21\right ) {\mathrm e}^{2 x}}{8}+\frac {\left (-x^{3}+3 c_{2} -6 x +6\right ) {\mathrm e}^{x}}{3}+\frac {5 x^{3}}{6}+\frac {15 x^{2}}{4}+\frac {35 x}{4}+c_{3} \]
✓ Solution by Mathematica
Time used: 0.885 (sec). Leaf size: 67
DSolve[y'''[x]-3*y''[x]+2*y'[x]==x^2*Exp[x]+3*x*Exp[2*x]+5*x^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {5 x^3}{6}+e^x \left (-\frac {x^3}{3}-2 x+c_1\right )+\frac {15 x^2}{4}+\frac {1}{8} e^{2 x} \left (6 x^2-18 x+21+4 c_2\right )+\frac {35 x}{4}+c_3 \]