19.46 problem section 9.3, problem 46

Internal problem ID [2193]
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 46
Date solved : Thursday, October 17, 2024 at 02:21:28 AM
CAS classification : [[_high_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} y^{\prime \prime \prime \prime }-8 y^{\prime \prime \prime }+32 y^{\prime \prime }-64 y^{\prime }+64 y&={\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \end{align*}

19.46.1 Solved as higher order constant coeff ode

Time used: 1.204 (sec)

The characteristic equation is

\[ \lambda ^{4}-8 \lambda ^{3}+32 \lambda ^{2}-64 \lambda +64 = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= 2-2 i\\ \lambda _2 &= 2+2 i\\ \lambda _3 &= 2-2 i\\ \lambda _4 &= 2+2 i \end{align*}

Therefore the homogeneous solution is

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

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

\begin{align*} y_1 &= {\mathrm e}^{\left (2+2 i\right ) x}\\ y_2 &= x \,{\mathrm e}^{\left (2+2 i\right ) x}\\ y_3 &= {\mathrm e}^{\left (2-2 i\right ) x}\\ y_4 &= x \,{\mathrm e}^{\left (2-2 i\right ) 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

\[ y^{\prime \prime \prime \prime }-8 y^{\prime \prime \prime }+32 y^{\prime \prime }-64 y^{\prime }+64 y = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime \prime }-8 y^{\prime \prime \prime }+32 y^{\prime \prime }-64 y^{\prime }+64 y = {\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \]

Let the particular solution be

\[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \]

Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows

\[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \]

Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coefficient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (x \right )\). This is given by

\begin{equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3&y_4\\ y_1'&y_2'&y_3'&y_4'\\ y_1''&y_2''&y_3''&y_4''\\ y_1'''&y_2'''&y_3'''&y_4'''\\ \end {vmatrix} \end{equation*}

Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives

\begin{align*} W &= \left [\begin {array}{cccc} {\mathrm e}^{\left (2+2 i\right ) x} & x \,{\mathrm e}^{\left (2+2 i\right ) x} & {\mathrm e}^{\left (2-2 i\right ) x} & x \,{\mathrm e}^{\left (2-2 i\right ) x} \\ \left (2+2 i\right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (1+\left (2+2 i\right ) x \right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (2-2 i\right ) {\mathrm e}^{\left (2-2 i\right ) x} & -2 \,{\mathrm e}^{\left (2-2 i\right ) x} \left (-\frac {1}{2}+\left (-1+i\right ) x \right ) \\ 8 i {\mathrm e}^{\left (2+2 i\right ) x} & \left (8 i x +4 i+4\right ) {\mathrm e}^{\left (2+2 i\right ) x} & -8 i {\mathrm e}^{\left (2-2 i\right ) x} & \left (-8 i x -4 i+4\right ) {\mathrm e}^{\left (2-2 i\right ) x} \\ \left (-16+16 i\right ) {\mathrm e}^{\left (2+2 i\right ) x} & 16 \,{\mathrm e}^{\left (2+2 i\right ) x} \left (\frac {3 i}{2}+\left (-1+i\right ) x \right ) & \left (-16-16 i\right ) {\mathrm e}^{\left (2-2 i\right ) x} & -16 \,{\mathrm e}^{\left (2-2 i\right ) x} \left (\frac {3 i}{2}+\left (1+i\right ) x \right ) \end {array}\right ] \\ |W| &= 256 \,{\mathrm e}^{\left (4+4 i\right ) x} {\mathrm e}^{\left (4-4 i\right ) x} \end{align*}

The determinant simplifies to

\begin{align*} |W| &= 256 \,{\mathrm e}^{8 x} \end{align*}

Now we determine \(W_i\) for each \(U_i\).

\begin{align*} W_1(x) &= \det \,\left [\begin {array}{ccc} x \,{\mathrm e}^{\left (2+2 i\right ) x} & {\mathrm e}^{\left (2-2 i\right ) x} & x \,{\mathrm e}^{\left (2-2 i\right ) x} \\ \left (1+\left (2+2 i\right ) x \right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (2-2 i\right ) {\mathrm e}^{\left (2-2 i\right ) x} & -2 \,{\mathrm e}^{\left (2-2 i\right ) x} \left (-\frac {1}{2}+\left (-1+i\right ) x \right ) \\ \left (8 i x +4 i+4\right ) {\mathrm e}^{\left (2+2 i\right ) x} & -8 i {\mathrm e}^{\left (2-2 i\right ) x} & \left (-8 i x -4 i+4\right ) {\mathrm e}^{\left (2-2 i\right ) x} \end {array}\right ] \\ &= 8 \,{\mathrm e}^{\left (6-2 i\right ) x} \left (i-2 x \right ) \end{align*}

\begin{align*} W_2(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (2+2 i\right ) x} & {\mathrm e}^{\left (2-2 i\right ) x} & x \,{\mathrm e}^{\left (2-2 i\right ) x} \\ \left (2+2 i\right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (2-2 i\right ) {\mathrm e}^{\left (2-2 i\right ) x} & -2 \,{\mathrm e}^{\left (2-2 i\right ) x} \left (-\frac {1}{2}+\left (-1+i\right ) x \right ) \\ 8 i {\mathrm e}^{\left (2+2 i\right ) x} & -8 i {\mathrm e}^{\left (2-2 i\right ) x} & \left (-8 i x -4 i+4\right ) {\mathrm e}^{\left (2-2 i\right ) x} \end {array}\right ] \\ &= -16 \,{\mathrm e}^{\left (6-2 i\right ) x} \end{align*}

\begin{align*} W_3(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (2+2 i\right ) x} & x \,{\mathrm e}^{\left (2+2 i\right ) x} & x \,{\mathrm e}^{\left (2-2 i\right ) x} \\ \left (2+2 i\right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (1+\left (2+2 i\right ) x \right ) {\mathrm e}^{\left (2+2 i\right ) x} & -2 \,{\mathrm e}^{\left (2-2 i\right ) x} \left (-\frac {1}{2}+\left (-1+i\right ) x \right ) \\ 8 i {\mathrm e}^{\left (2+2 i\right ) x} & \left (8 i x +4 i+4\right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (-8 i x -4 i+4\right ) {\mathrm e}^{\left (2-2 i\right ) x} \end {array}\right ] \\ &= -8 \,{\mathrm e}^{\left (6+2 i\right ) x} \left (i+2 x \right ) \end{align*}

\begin{align*} W_4(x) &= \det \,\left [\begin {array}{ccc} {\mathrm e}^{\left (2+2 i\right ) x} & x \,{\mathrm e}^{\left (2+2 i\right ) x} & {\mathrm e}^{\left (2-2 i\right ) x} \\ \left (2+2 i\right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (1+\left (2+2 i\right ) x \right ) {\mathrm e}^{\left (2+2 i\right ) x} & \left (2-2 i\right ) {\mathrm e}^{\left (2-2 i\right ) x} \\ 8 i {\mathrm e}^{\left (2+2 i\right ) x} & \left (8 i x +4 i+4\right ) {\mathrm e}^{\left (2+2 i\right ) x} & -8 i {\mathrm e}^{\left (2-2 i\right ) x} \end {array}\right ] \\ &= -16 \,{\mathrm e}^{\left (6+2 i\right ) x} \end{align*}

Now we are ready to evaluate each \(U_i(x)\).

\begin{align*} U_1 &= (-1)^{4-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{3} \int { \frac { \left ({\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right )\right ) \left (8 \,{\mathrm e}^{\left (6-2 i\right ) x} \left (i-2 x \right )\right )}{\left (1\right ) \left (256 \,{\mathrm e}^{8 x}\right )} \, dx} \\ &= - \int { \frac {8 \,{\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{\left (6-2 i\right ) x} \left (i-2 x \right )}{256 \,{\mathrm e}^{8 x}} \, dx}\\ &= - \int {\left (\frac {\left (i-2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{32}\right ) \, dx} \\ &= -\left (\int \frac {\left (i-2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{32}d x \right ) \\ &= -\left (\int \frac {\left (i-2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{32}d x \right ) \end{align*}

\begin{align*} U_2 &= (-1)^{4-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left ({\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right )\right ) \left (-16 \,{\mathrm e}^{\left (6-2 i\right ) x}\right )}{\left (1\right ) \left (256 \,{\mathrm e}^{8 x}\right )} \, dx} \\ &= \int { \frac {-16 \,{\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{\left (6-2 i\right ) x}}{256 \,{\mathrm e}^{8 x}} \, dx}\\ &= \int {\left (-\frac {\left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{16}\right ) \, dx} \\ &= -\frac {x}{32}-\frac {i x}{32}-\frac {{\mathrm e}^{-4 i x}}{128}-\frac {i {\mathrm e}^{-4 i x}}{128} \\ &= -\frac {x}{32}-\frac {i x}{32}-\frac {{\mathrm e}^{-4 i x}}{128}-\frac {i {\mathrm e}^{-4 i x}}{128} \end{align*}

\begin{align*} U_3 &= (-1)^{4-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left ({\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right )\right ) \left (-8 \,{\mathrm e}^{\left (6+2 i\right ) x} \left (i+2 x \right )\right )}{\left (1\right ) \left (256 \,{\mathrm e}^{8 x}\right )} \, dx} \\ &= - \int { \frac {-8 \,{\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{\left (6+2 i\right ) x} \left (i+2 x \right )}{256 \,{\mathrm e}^{8 x}} \, dx}\\ &= - \int {\left (-\frac {\left (i+2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{32}\right ) \, dx} \\ &= -\left (\int -\frac {\left (i+2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{32}d x \right ) \\ &= -\left (\int -\frac {\left (i+2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{32}d x \right ) \end{align*}

\begin{align*} U_4 &= (-1)^{4-4} \int { \frac {F(x) W_4(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left ({\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right )\right ) \left (-16 \,{\mathrm e}^{\left (6+2 i\right ) x}\right )}{\left (1\right ) \left (256 \,{\mathrm e}^{8 x}\right )} \, dx} \\ &= \int { \frac {-16 \,{\mathrm e}^{2 x} \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{\left (6+2 i\right ) x}}{256 \,{\mathrm e}^{8 x}} \, dx}\\ &= \int {\left (-\frac {\left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{16}\right ) \, dx} \\ &= -\frac {x}{32}+\frac {i x}{32}-\frac {{\mathrm e}^{4 i x}}{128}+\frac {i {\mathrm e}^{4 i x}}{128} \\ &= -\frac {x}{32}+\frac {i x}{32}-\frac {{\mathrm e}^{4 i x}}{128}+\frac {i {\mathrm e}^{4 i x}}{128} \end{align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from

\[ y_p = U_1 y_1+U_2 y_2+U_3 y_3+U_4 y_4 \]

Hence

\begin{equation*} \begin {split} y_p &= \left (-\left (\int \frac {\left (i-2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{32}d x \right )\right ) \left ({\mathrm e}^{\left (2+2 i\right ) x}\right ) \\ &+\left (-\frac {x}{32}-\frac {i x}{32}-\frac {{\mathrm e}^{-4 i x}}{128}-\frac {i {\mathrm e}^{-4 i x}}{128}\right ) \left (x \,{\mathrm e}^{\left (2+2 i\right ) x}\right ) \\ &+\left (-\left (\int -\frac {\left (i+2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{32}d x \right )\right ) \left ({\mathrm e}^{\left (2-2 i\right ) x}\right ) \\ &+\left (-\frac {x}{32}+\frac {i x}{32}-\frac {{\mathrm e}^{4 i x}}{128}+\frac {i {\mathrm e}^{4 i x}}{128}\right ) \left (x \,{\mathrm e}^{\left (2-2 i\right ) x}\right ) \end {split} \end{equation*}

Therefore the particular solution is

\[ y_p = -\frac {\left (\int \left (i-2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{-2 i x}d x \right ) {\mathrm e}^{\left (2+2 i\right ) x}}{32}+\frac {\left (\int \left (i+2 x \right ) \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) {\mathrm e}^{2 i x}d x \right ) {\mathrm e}^{\left (2-2 i\right ) x}}{32}+\frac {\left (\left (-\frac {1}{4}-\frac {i}{4}+\left (-1+i\right ) x \right ) {\mathrm e}^{\left (2-2 i\right ) x}-\left (\frac {1}{4}-\frac {i}{4}+\left (1+i\right ) x \right ) {\mathrm e}^{\left (2+2 i\right ) x}\right ) x}{32} \]

Which simplifies to

\[ y_p = -\frac {{\mathrm e}^{2 x} \left (-\left (\int \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \left (2 \cos \left (2 x \right ) x -\sin \left (2 x \right )\right )d x \right ) \cos \left (2 x \right )-\left (\int \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \left (\cos \left (2 x \right )+2 \sin \left (2 x \right ) x \right )d x \right ) \sin \left (2 x \right )+x \left (\left (\frac {1}{4}+x \right ) \cos \left (2 x \right )-\left (-\frac {1}{4}+x \right ) \sin \left (2 x \right )\right )\right )}{16} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\left (2+2 i\right ) x} c_1 +x \,{\mathrm e}^{\left (2+2 i\right ) x} c_2 +{\mathrm e}^{\left (2-2 i\right ) x} c_3 +x \,{\mathrm e}^{\left (2-2 i\right ) x} c_4\right ) + \left (-\frac {{\mathrm e}^{2 x} \left (-\left (\int \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \left (2 \cos \left (2 x \right ) x -\sin \left (2 x \right )\right )d x \right ) \cos \left (2 x \right )-\left (\int \left (\cos \left (2 x \right )-\sin \left (2 x \right )\right ) \left (\cos \left (2 x \right )+2 \sin \left (2 x \right ) x \right )d x \right ) \sin \left (2 x \right )+x \left (\left (\frac {1}{4}+x \right ) \cos \left (2 x \right )-\left (-\frac {1}{4}+x \right ) \sin \left (2 x \right )\right )\right )}{16}\right ) \\ \end{align*}

19.46.2 Maple step by step solution

19.46.3 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`
 

19.46.4 Maple dsolve solution

Solving time : 0.233 (sec)
Leaf size : 52

dsolve(diff(diff(diff(diff(y(x),x),x),x),x)-8*diff(diff(diff(y(x),x),x),x)+32*diff(diff(y(x),x),x)-64*diff(y(x),x)+64*y(x) = exp(2*x)*(cos(2*x)-sin(2*x)), 
       y(x),singsol=all)
 

\[ y = -\frac {\left (\left (x^{2}+\left (-32 c_4 -\frac {3}{2}\right ) x -32 c_1 -\frac {41}{4}\right ) \cos \left (2 x \right )-\sin \left (2 x \right ) \left (x^{2}+\left (32 c_3 +\frac {1}{2}\right ) x +32 c_2 -\frac {517}{18}\right )\right ) {\mathrm e}^{2 x}}{32} \]

19.46.5 Mathematica DSolve solution

Solving time : 0.172 (sec)
Leaf size : 65

DSolve[{1*D[y[x],{x,4}]-8*D[y[x],{x,3}]+32*D[y[x],{x,2}]-64*D[y[x],x]+64*y[x]==Exp[2*x]*(Cos[2*x]-Sin[2*x]),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \frac {1}{256} e^{2 x} \left (\left (-8 x^2+4 (1+64 c_4) x+5+256 c_3\right ) \cos (2 x)+\left (8 x^2+8 (1+32 c_2) x-1+256 c_1\right ) \sin (2 x)\right ) \]