problem 2

Internal problem ID [12774]
Internal ﬁle name [OUTPUT/11427_Friday_November_03_2023_06_32_50_AM_27271063/index.tex]

Book: Ordinary Diﬀerential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Diﬀerential Equations. Exercises 4.4, page 218
Problem number: 2.
ODE order: 4.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime \prime \prime }+4 y^{\prime \prime }=24 x^{2}-6 x +14+32 \cos \left (2 x \right )} \] 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 \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{4}+4 \lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 0\\ \lambda _3 &= 2 i\\ \lambda _4 &= -2 i \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= 1 \\ y_2 &= x \\ y_3 &= {\mathrm e}^{-2 i x} \\ y_4 &= {\mathrm e}^{2 i x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime \prime }+4 y^{\prime \prime } = 24 x^{2}-6 x +14+32 \cos \left (2 x \right ) \] 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} 1 & x & {\mathrm e}^{-2 i x} & {\mathrm e}^{2 i x} \\ 0 & 1 & -2 i {\mathrm e}^{-2 i x} & 2 i {\mathrm e}^{2 i x} \\ 0 & 0 & -4 \,{\mathrm e}^{-2 i x} & -4 \,{\mathrm e}^{2 i x} \\ 0 & 0 & 8 i {\mathrm e}^{-2 i x} & -8 i {\mathrm e}^{2 i x} \end {array}\right ] \\ |W| &= 64 i {\mathrm e}^{-2 i x} {\mathrm e}^{2 i 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}^{-2 i x} & {\mathrm e}^{2 i x} \\ 1 & -2 i {\mathrm e}^{-2 i x} & 2 i {\mathrm e}^{2 i x} \\ 0 & -4 \,{\mathrm e}^{-2 i x} & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= 16 i x \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{ccc} 1 & {\mathrm e}^{-2 i x} & {\mathrm e}^{2 i x} \\ 0 & -2 i {\mathrm e}^{-2 i x} & 2 i {\mathrm e}^{2 i x} \\ 0 & -4 \,{\mathrm e}^{-2 i x} & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= 16 i \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{ccc} 1 & x & {\mathrm e}^{2 i x} \\ 0 & 1 & 2 i {\mathrm e}^{2 i x} \\ 0 & 0 & -4 \,{\mathrm e}^{2 i x} \end {array}\right ] \\ &= -4 \,{\mathrm e}^{2 i x} \end {align*}

\begin {align*} W_4(x) &= \det \,\left [\begin {array}{ccc} 1 & x & {\mathrm e}^{-2 i x} \\ 0 & 1 & -2 i {\mathrm e}^{-2 i x} \\ 0 & 0 & -4 \,{\mathrm e}^{-2 i x} \end {array}\right ] \\ &= -4 \,{\mathrm e}^{-2 i 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 (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) \left (16 i x\right )}{\left (1\right ) \left (64 i\right )} \, dx} \\ &= - \int { \frac {16 i \left (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) x}{64 i} \, dx}\\ &= - \int {\left (\frac {\left (7+12 x^{2}-3 x +16 \cos \left (2 x \right )\right ) x}{2}\right ) \, dx} \\ &= -\frac {3 x^{4}}{2}-2 \cos \left (2 x \right )-4 x \sin \left (2 x \right )+\frac {x^{3}}{2}-\frac {7 x^{2}}{4} \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 (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) \left (16 i\right )}{\left (1\right ) \left (64 i\right )} \, dx} \\ &= \int { \frac {16 i \left (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right )}{64 i} \, dx}\\ &= \int {\left (6 x^{2}-\frac {3 x}{2}+\frac {7}{2}+8 \cos \left (2 x \right )\right ) \, dx}\\ &= \frac {7 x}{2}-\frac {3 x^{2}}{4}+2 x^{3}+4 \sin \left (2 x \right ) \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 (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) \left (-4 \,{\mathrm e}^{2 i x}\right )}{\left (1\right ) \left (64 i\right )} \, dx} \\ &= - \int { \frac {-4 \left (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{64 i} \, dx}\\ &= - \int {\left (\frac {i \left (7+12 x^{2}-3 x +16 \cos \left (2 x \right )\right ) {\mathrm e}^{2 i x}}{8}\right ) \, dx} \\ &= -i x -\frac {{\mathrm e}^{4 i x}}{4}-\frac {\left (24 x^{2}+24 i x -6 x -3 i+2\right ) {\mathrm e}^{2 i x}}{32} \\ &= -i x -\frac {{\mathrm e}^{4 i x}}{4}-\frac {\left (24 x^{2}+24 i x -6 x -3 i+2\right ) {\mathrm e}^{2 i x}}{32} \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 (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) \left (-4 \,{\mathrm e}^{-2 i x}\right )}{\left (1\right ) \left (64 i\right )} \, dx} \\ &= \int { \frac {-4 \left (24 x^{2}-6 x +14+32 \cos \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{64 i} \, dx}\\ &= \int {\left (\frac {i \left (7+12 x^{2}-3 x +16 \cos \left (2 x \right )\right ) {\mathrm e}^{-2 i x}}{8}\right ) \, dx} \\ &= -\frac {\left (24 x^{2}-24 i x -6 x +3 i+2\right ) {\mathrm e}^{-2 i x}}{32}+i x -\frac {{\mathrm e}^{-4 i x}}{4} \\ &= -\frac {\left (24 x^{2}-24 i x -6 x +3 i+2\right ) {\mathrm e}^{-2 i x}}{32}+i x -\frac {{\mathrm e}^{-4 i x}}{4} \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 (-\frac {3 x^{4}}{2}-2 \cos \left (2 x \right )-4 x \sin \left (2 x \right )+\frac {x^{3}}{2}-\frac {7 x^{2}}{4}\right ) \\ &+\left (\frac {7 x}{2}-\frac {3 x^{2}}{4}+2 x^{3}+4 \sin \left (2 x \right )\right ) \left (x\right ) \\ &+\left (-i x -\frac {{\mathrm e}^{4 i x}}{4}-\frac {\left (24 x^{2}+24 i x -6 x -3 i+2\right ) {\mathrm e}^{2 i x}}{32}\right ) \left ({\mathrm e}^{-2 i x}\right ) \\ &+\left (-\frac {\left (24 x^{2}-24 i x -6 x +3 i+2\right ) {\mathrm e}^{-2 i x}}{32}+i x -\frac {{\mathrm e}^{-4 i x}}{4}\right ) \left ({\mathrm e}^{2 i x}\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = \frac {x^{2}}{4}+\frac {x^{4}}{2}+\frac {3 x}{8}-\frac {x^{3}}{4}-\frac {1}{8}-\frac {5 \cos \left (2 x \right )}{2}-2 x \sin \left (2 x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{2} x +c_{1} +{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4}\right ) + \left (\frac {x^{2}}{4}+\frac {x^{4}}{2}+\frac {3 x}{8}-\frac {x^{3}}{4}-\frac {1}{8}-\frac {5 \cos \left (2 x \right )}{2}-2 x \sin \left (2 x \right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} x +c_{1} +{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} +\frac {x^{2}}{4}+\frac {x^{4}}{2}+\frac {3 x}{8}-\frac {x^{3}}{4}-\frac {1}{8}-\frac {5 \cos \left (2 x \right )}{2}-2 x \sin \left (2 x \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{2} x +c_{1} +{\mathrm e}^{-2 i x} c_{3} +{\mathrm e}^{2 i x} c_{4} +\frac {x^{2}}{4}+\frac {x^{4}}{2}+\frac {3 x}{8}-\frac {x^{3}}{4}-\frac {1}{8}-\frac {5 \cos \left (2 x \right )}{2}-2 x \sin \left (2 x \right ) \] Veriﬁed OK.

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] 
-> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = 24*_a^2+32*cos(2*_a)-4*_b(_a)-6*_a+14, _b(_a)`   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying high order exact linear fully integrable 
   trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
   trying a double symmetry of the form [xi=0, eta=F(x)] 
   -> Try solving first the homogeneous part of the ODE 
      checking if the LODE has constant coefficients 
      <- constant coefficients successful 
   <- solving first the homogeneous part of the ODE successful 
<- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful`

\[ y \left (x \right ) = \frac {\left (-c_{1} -10\right ) \cos \left (2 x \right )}{4}+\frac {\left (-8 x -c_{2} \right ) \sin \left (2 x \right )}{4}+\frac {x^{4}}{2}-\frac {x^{3}}{4}+\frac {x^{2}}{4}+c_{3} x +c_{4} \]

\[ y(x)\to \frac {1}{4} \left (2 x^4-x^3+x^2+4 c_4 x-(12+c_1) \cos (2 x)-(8 x+c_2) \sin (2 x)+4 c_3\right ) \]

11.2 problem 2