Internal problem ID [1509]
Internal file name [OUTPUT/1510_Sunday_June_05_2022_02_20_06_AM_92045200/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 12.
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]]
\[ \boxed {8 y^{\prime \prime \prime }-12 y^{\prime \prime }+6 y^{\prime }-y={\mathrm e}^{\frac {x}{2}} \left (4 x +1\right )} \] 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 \[ 8 y^{\prime \prime \prime }-12 y^{\prime \prime }+6 y^{\prime }-y = 0 \] The characteristic equation is \[ 8 \lambda ^{3}-12 \lambda ^{2}+6 \lambda -1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= {\frac {1}{2}}\\ \lambda _2 &= {\frac {1}{2}}\\ \lambda _3 &= {\frac {1}{2}} \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{\frac {x}{2}} x +x^{2} {\mathrm e}^{\frac {x}{2}} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{\frac {x}{2}} \\ y_2 &= {\mathrm e}^{\frac {x}{2}} x \\ y_3 &= x^{2} {\mathrm e}^{\frac {x}{2}} \\ \end{align*} Now the particular solution to the given ODE is found \[ 8 y^{\prime \prime \prime }-12 y^{\prime \prime }+6 y^{\prime }-y = {\mathrm e}^{\frac {x}{2}} \left (4 x +1\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{\frac {x}{2}} \left (4 x +1\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ \left [\left \{{\mathrm e}^{\frac {x}{2}} x, {\mathrm e}^{\frac {x}{2}}\right \}\right ] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \left \{x^{2} {\mathrm e}^{\frac {x}{2}}, {\mathrm e}^{\frac {x}{2}} x, {\mathrm e}^{\frac {x}{2}}\right \} \] Since \({\mathrm e}^{\frac {x}{2}}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ \left [\left \{x^{2} {\mathrm e}^{\frac {x}{2}}, {\mathrm e}^{\frac {x}{2}} x\right \}\right ] \] Since \({\mathrm e}^{\frac {x}{2}} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ \left [\left \{x^{2} {\mathrm e}^{\frac {x}{2}}, x^{3} {\mathrm e}^{\frac {x}{2}}\right \}\right ] \] Since \(x^{2} {\mathrm e}^{\frac {x}{2}}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ \left [\left \{x^{3} {\mathrm e}^{\frac {x}{2}}, x^{4} {\mathrm e}^{\frac {x}{2}}\right \}\right ] \] 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. \[ y_p = A_{1} x^{3} {\mathrm e}^{\frac {x}{2}}+A_{2} x^{4} {\mathrm e}^{\frac {x}{2}} \] 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 \[ 48 A_{1} {\mathrm e}^{\frac {x}{2}}+192 A_{2} x \,{\mathrm e}^{\frac {x}{2}} = {\mathrm e}^{\frac {x}{2}} \left (4 x +1\right ) \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{48}}, A_{2} = {\frac {1}{48}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{3} {\mathrm e}^{\frac {x}{2}}}{48}+\frac {x^{4} {\mathrm e}^{\frac {x}{2}}}{48} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{\frac {x}{2}} x +x^{2} {\mathrm e}^{\frac {x}{2}} c_{3}\right ) + \left (\frac {x^{3} {\mathrm e}^{\frac {x}{2}}}{48}+\frac {x^{4} {\mathrm e}^{\frac {x}{2}}}{48}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{\frac {x}{2}} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{\frac {x}{2}}}{48}+\frac {x^{4} {\mathrm e}^{\frac {x}{2}}}{48} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\frac {x}{2}} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{\frac {x}{2}}}{48}+\frac {x^{4} {\mathrm e}^{\frac {x}{2}}}{48} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\frac {x}{2}} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{\frac {x}{2}}}{48}+\frac {x^{4} {\mathrm e}^{\frac {x}{2}}}{48} \] Verified OK.
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: 0.016 (sec). Leaf size: 33
dsolve(8*diff(y(x),x$3)-12*diff(y(x),x$2)+6*diff(y(x),x)-y(x)=exp(x/2)*(1+4*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (x^{4}+x^{3}+\left (48 c_{2} +\frac {3}{16}\right ) x^{2}+48 c_{3} x +48 c_{1} \right ) {\mathrm e}^{\frac {x}{2}}}{48} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 39
DSolve[8*y'''[x]-12*y''[x]+6*y'[x]-y[x]==Exp[x/2]*(1+4*x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{48} e^{x/2} \left (x^4+x^3+48 c_3 x^2+48 c_2 x+48 c_1\right ) \]