problem 25

Internal problem ID [5432]
Internal ﬁle name [OUTPUT/4923_Tuesday_February_06_2024_10_14_19_PM_17103159/index.tex]

Book: Schaums Outline. Theory and problems of Diﬀerential Equations, 1st edition. Frank Ayres. McGraw Hill 1952
Section: Chapter 19. Linear equations with variable coeﬃcients (Misc. types). Supplemetary problems. Page 132
Problem number: 25.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }=x^{2}} \] 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 \[ y^{\prime \prime \prime }+y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= x \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime } = x^{2} \] The particular solution is found using the method of undetermined coeﬃcients. Looking at the RHS of the ode, which is \[ x^{2} \] 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 \[ \{1, x, {\mathrm e}^{-x}\} \] 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 \[ 12 x^{2} A_{3}+6 x A_{2}+24 x A_{3}+2 A_{1}+6 A_{2} = x^{2} \] Solving for the unknowns by comparing coeﬃcients results in \[ \left [A_{1} = 1, A_{2} = -{\frac {1}{3}}, A_{3} = {\frac {1}{12}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {1}{12} x^{4}-\frac {1}{3} x^{3}+x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x\right ) + \left (\frac {1}{12} x^{4}-\frac {1}{3} x^{3}+x^{2}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +\frac {x^{4}}{12}-\frac {x^{3}}{3}+x^{2} \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x +\frac {x^{4}}{12}-\frac {x^{3}}{3}+x^{2} \] 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-_b(_a), _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`

\[ y \left (x \right ) = \frac {x^{4}}{12}+x^{2}-\frac {x^{3}}{3}+c_{1} {\mathrm e}^{-x}+c_{2} x +c_{3} \]

14.4 problem 25