Internal problem ID [2153]
Internal file name [OUTPUT/2153_Monday_February_26_2024_09_17_54_AM_42651828/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 19, page 86
Problem number: 14.
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]]
\[ \boxed {y^{\prime \prime \prime }-4 y^{\prime \prime }=x^{2}+8} \] 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 \[ y^{\prime \prime \prime }-4 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-4 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 4\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{4 x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-4 y^{\prime \prime } = x^{2}+8 \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x^{2}+1 \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x, x^{2}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, x, {\mathrm e}^{4 x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}, x^{3}\}] \] Since \(x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2}, x^{3}, x^{4}\}] \] 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. \[ y_p = A_{3} x^{4}+A_{2} x^{3}+A_{1} x^{2} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) 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 \[ -48 x^{2} A_{3}-24 x A_{2}+24 x A_{3}-8 A_{1}+6 A_{2} = x^{2}+8 \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = -{\frac {65}{64}}, A_{2} = -{\frac {1}{48}}, A_{3} = -{\frac {1}{48}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = -\frac {1}{48} x^{4}-\frac {1}{48} x^{3}-\frac {65}{64} x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3}\right ) + \left (-\frac {1}{48} x^{4}-\frac {1}{48} x^{3}-\frac {65}{64} x^{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} -\frac {x^{4}}{48}-\frac {x^{3}}{48}-\frac {65 x^{2}}{64} \\ \end{align*}
Verification of solutions
\[ y = c_{2} x +c_{1} +{\mathrm e}^{4 x} c_{3} -\frac {x^{4}}{48}-\frac {x^{3}}{48}-\frac {65 x^{2}}{64} \] Verified OK.
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable -> Calling odsolve with the ODE`, diff(_b(_a), _a) = _a^2+4*_b(_a)+8, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 31
dsolve(diff(y(x),x$3)-4*diff(y(x),x$2)=x^2+8,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {65 x^{2}}{64}-\frac {x^{3}}{48}-\frac {x^{4}}{48}+\frac {{\mathrm e}^{4 x} c_{1}}{16}+c_{2} x +c_{3} \]
✓ Solution by Mathematica
Time used: 0.143 (sec). Leaf size: 41
DSolve[y'''[x]-4*y''[x]==x^2+8,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{192} \left (-4 x^4-4 x^3-195 x^2+12 c_1 e^{4 x}\right )+c_3 x+c_2 \]