19.25 problem section 9.3, problem 25

Internal problem ID [2172]
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 25
Date solved : Sunday, October 06, 2024 at 03:12:10 AM
CAS classification : [[_high_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+2 y^{\prime }-y&={\mathrm e}^{x} \left (1+x \right ) \end{align*}

19.25.1 Solved as higher order constant coeff ode

Time used: 0.127 (sec)

The characteristic equation is

\[ \lambda ^{4}-2 \lambda ^{3}+2 \lambda -1 = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= -1\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end{align*}

Therefore the homogeneous solution is

\[ y_h(x)={\mathrm e}^{-x} c_1 +{\mathrm e}^{x} c_2 +x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4 \]

The fundamental set of solutions for the homogeneous solution are the following

\begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= {\mathrm e}^{x} x\\ y_4 &= x^{2} {\mathrm e}^{x} \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

\[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+2 y^{\prime }-y = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime \prime }-2 y^{\prime \prime \prime }+2 y^{\prime }-y = {\mathrm e}^{x} \left (1+x \right ) \]

The particular solution is now found using the method of undetermined coefficients.

Looking at the RHS of the ode, which is

\[ {\mathrm e}^{x} \left (1+x \right ) \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x}\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x, {\mathrm e}^{x}, {\mathrm e}^{-x}\} \]

Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x^{2} {\mathrm e}^{x}, {\mathrm e}^{x} x\}] \]

Since \({\mathrm e}^{x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x^{2} {\mathrm e}^{x}, x^{3} {\mathrm e}^{x}\}] \]

Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x^{3} {\mathrm e}^{x}, x^{4} {\mathrm e}^{x}\}] \]

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.

\[ y_p = A_{1} x^{3} {\mathrm e}^{x}+A_{2} x^{4} {\mathrm e}^{x} \]

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

\[ 12 A_{1} {\mathrm e}^{x}+24 A_{2} {\mathrm e}^{x}+48 A_{2} x \,{\mathrm e}^{x} = {\mathrm e}^{x} \left (1+x \right ) \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{24}}, A_{2} = {\frac {1}{48}}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {x^{3} {\mathrm e}^{x}}{24}+\frac {x^{4} {\mathrm e}^{x}}{48} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_1 +{\mathrm e}^{x} c_2 +x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4\right ) + \left (\frac {x^{3} {\mathrm e}^{x}}{24}+\frac {x^{4} {\mathrm e}^{x}}{48}\right ) \\ \end{align*}

19.25.2 Maple step by step solution

19.25.3 Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 4; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 4; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

19.25.4 Maple dsolve solution

Solving time : 0.008 (sec)
Leaf size : 37

dsolve(diff(diff(diff(diff(y(x),x),x),x),x)-2*diff(diff(diff(y(x),x),x),x)+2*diff(y(x),x)-y(x) = exp(x)*(x+1), 
       y(x),singsol=all)
 

\[ y = c_2 \,{\mathrm e}^{-x}+\frac {{\mathrm e}^{x} \left (x^{4}+48 c_4 \,x^{2}+2 x^{3}+48 x c_3 +48 c_1 \right )}{48} \]

19.25.5 Mathematica DSolve solution

Solving time : 0.026 (sec)
Leaf size : 55

DSolve[{1*D[y[x],{x,4}]-2*D[y[x],{x,3}]-0*D[y[x],{x,2}]+2*D[y[x],x]-1*y[x]==Exp[x]*(1+x),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \frac {1}{96} e^x \left (2 x^4+4 x^3+(-6+96 c_4) x^2+(6+96 c_3) x-3+96 c_2\right )+c_1 e^{-x} \]