2.8.12 problem 15

Internal problem ID [18267]
Book : Elementary Differential Equations. By Thornton C. Fry. D Van Nostrand. NY. First Edition (1929)
Section : Chapter VII. Linear equations of order higher than the first. section 56. Problems at page 163
Problem number : 15
Date solved : Monday, December 23, 2024 at 09:47:15 PM
CAS classification : [[_high_order, _exact, _linear, _nonhomogeneous]]

Solve

\begin{align*} t^{4} x^{\prime \prime \prime \prime }-2 t^{3} x^{\prime \prime \prime }-20 t^{2} x^{\prime \prime }+12 x^{\prime } t +16 x&=\cos \left (3 \ln \left (t \right )\right ) \end{align*}

Solved as higher order Euler type ode

Time used: 0.786 (sec)

This is Euler ODE of higher order. Let \(x = t^{\lambda }\). Hence

\begin{align*} x^{\prime } &= \lambda \,t^{\lambda -1}\\ x^{\prime \prime } &= \lambda \left (\lambda -1\right ) t^{\lambda -2}\\ x^{\prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) t^{\lambda -3}\\ x^{\prime \prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) t^{\lambda -4} \end{align*}

Substituting these back into

\[ t^{4} x^{\prime \prime \prime \prime }-2 t^{3} x^{\prime \prime \prime }-20 t^{2} x^{\prime \prime }+12 x^{\prime } t +16 x = \cos \left (3 \ln \left (t \right )\right ) \]

gives

\[ 12 t \lambda \,t^{\lambda -1}-20 t^{2} \lambda \left (\lambda -1\right ) t^{\lambda -2}-2 t^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) t^{\lambda -3}+t^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) t^{\lambda -4}+16 t^{\lambda } = 0 \]

Which simplifies to

\[ 12 \lambda \,t^{\lambda }-20 \lambda \left (\lambda -1\right ) t^{\lambda }-2 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) t^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) t^{\lambda }+16 t^{\lambda } = 0 \]

And since \(t^{\lambda }\neq 0\) then dividing through by \(t^{\lambda }\), the above becomes

\[ 12 \lambda -20 \lambda \left (\lambda -1\right )-2 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )+16 = 0 \]

Simplifying gives the characteristic equation as

\[ \left (\lambda -2\right ) \left (\lambda -8\right ) \left (\lambda +1\right )^{2} = 0 \]

Solving the above gives the following roots

\begin{align*} \lambda _1 &= 2\\ \lambda _2 &= 8\\ \lambda _3 &= -1\\ \lambda _4 &= -1 \end{align*}

This table summarises the result

root	multiplicity	type of root
\(-1\)	\(2\)	real root
\(2\)	\(1\)	real root
\(8\)	\(1\)	real root

The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1t^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1t^{\lambda }\) and \(c_2t^{\lambda } \ln \left (t \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1t^{\lambda }\) and \(c_2t^{\lambda } \ln \left (t \right )\) and \(c_3t^{\lambda } \ln \left (t \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(t^{\alpha } \left (c_1\cos (\beta \ln \left (t \right ))+c_2\sin (\beta \ln \left (t \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (t \right ) t^{\alpha }\left (c_1\cos (\beta \ln \left (t \right ))+c_2\sin (\beta \ln \left (t \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (t \right )^{2} t^{\alpha }\left (c_1\cos (\beta \ln \left (t \right ))+c_2\sin (\beta \ln \left (t \right ))\right )\) basis solutions. And so on. Using the above show that the solution is

\[ x = \frac {c_1}{t}+\frac {c_2 \ln \left (t \right )}{t}+c_3 \,t^{2}+c_4 \,t^{8} \]

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

\begin{align*} x_1 &= \frac {1}{t} \\ x_2 &= \frac {\ln \left (t \right )}{t} \\ x_3 &= t^{2} \\ x_4 &= t^{8} \\ \end{align*}

This is higher order nonhomogeneous Euler type ODE. Let the solution be

\[ x = x_h + x_p \]

Where \(x_h\) is the solution to the homogeneous Euler ODE And \(x_p\) is a particular solution to the nonhomogeneous Euler ODE. \(x_h\) is the solution to

\[ t^{4} x^{\prime \prime \prime \prime }-2 t^{3} x^{\prime \prime \prime }-20 t^{2} x^{\prime \prime }+12 x^{\prime } t +16 x = 0 \]

Now the particular solution to the given ODE is found

\[ t^{4} x^{\prime \prime \prime \prime }-2 t^{3} x^{\prime \prime \prime }-20 t^{2} x^{\prime \prime }+12 x^{\prime } t +16 x = \cos \left (3 \ln \left (t \right )\right ) \]

Let the particular solution be

\[ x_p = U_1 x_1+U_2 x_2+U_3 x_3+U_4 x_4 \]

Where \(x_i\) are the basis solutions found above for the homogeneous solution \(x_h\) and \(U_i(t)\) are functions to be determined as follows

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

Where \(W(t)\) is the Wronskian and \(W_i(t)\) 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 \(F(t)\) is the RHS of the ODE. Therefore, the first step is to find the Wronskian \(W \left (t \right )\). This is given by

\begin{equation*} W(t) = \begin {vmatrix} x_1&x_2&x_3&x_4\\ x_1'&x_2'&x_3'&x_4'\\ x_1''&x_2''&x_3''&x_4''\\ x_1'''&x_2'''&x_3'''&x_4'''\\ \end {vmatrix} \end{equation*}

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

\begin{align*} W &= \left [\begin {array}{cccc} \frac {1}{t} & \frac {\ln \left (t \right )}{t} & t^{2} & t^{8} \\ -\frac {1}{t^{2}} & \frac {1-\ln \left (t \right )}{t^{2}} & 2 t & 8 t^{7} \\ \frac {2}{t^{3}} & \frac {-3+2 \ln \left (t \right )}{t^{3}} & 2 & 56 t^{6} \\ -\frac {6}{t^{4}} & \frac {11-6 \ln \left (t \right )}{t^{4}} & 0 & 336 t^{5} \end {array}\right ] \\ |W| &= 4374 t^{2} \end{align*}

The determinant simplifies to

\begin{align*} |W| &= 4374 t^{2} \end{align*}

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

\begin{align*} W_1(t) &= \det \,\left [\begin {array}{ccc} \frac {\ln \left (t \right )}{t} & t^{2} & t^{8} \\ \frac {1-\ln \left (t \right )}{t^{2}} & 2 t & 8 t^{7} \\ \frac {-3+2 \ln \left (t \right )}{t^{3}} & 2 & 56 t^{6} \end {array}\right ] \\ &= 18 t^{6} \left (-4+9 \ln \left (t \right )\right ) \end{align*}

\begin{align*} W_2(t) &= \det \,\left [\begin {array}{ccc} \frac {1}{t} & t^{2} & t^{8} \\ -\frac {1}{t^{2}} & 2 t & 8 t^{7} \\ \frac {2}{t^{3}} & 2 & 56 t^{6} \end {array}\right ] \\ &= 162 t^{6} \end{align*}

\begin{align*} W_3(t) &= \det \,\left [\begin {array}{ccc} \frac {1}{t} & \frac {\ln \left (t \right )}{t} & t^{8} \\ -\frac {1}{t^{2}} & \frac {1-\ln \left (t \right )}{t^{2}} & 8 t^{7} \\ \frac {2}{t^{3}} & \frac {-3+2 \ln \left (t \right )}{t^{3}} & 56 t^{6} \end {array}\right ] \\ &= 81 t^{3} \end{align*}

\begin{align*} W_4(t) &= \det \,\left [\begin {array}{ccc} \frac {1}{t} & \frac {\ln \left (t \right )}{t} & t^{2} \\ -\frac {1}{t^{2}} & \frac {1-\ln \left (t \right )}{t^{2}} & 2 t \\ \frac {2}{t^{3}} & \frac {-3+2 \ln \left (t \right )}{t^{3}} & 2 \end {array}\right ] \\ &= \frac {9}{t^{3}} \end{align*}

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

\begin{align*} U_1 &= (-1)^{4-1} \int { \frac {F(t) W_1(t) }{a W(t)} \, dt}\\ &= (-1)^{3} \int { \frac { \left (\cos \left (3 \ln \left (t \right )\right )\right ) \left (18 t^{6} \left (-4+9 \ln \left (t \right )\right )\right )}{\left (t^{4}\right ) \left (4374 t^{2}\right )} \, dt} \\ &= - \int { \frac {18 \cos \left (3 \ln \left (t \right )\right ) t^{6} \left (-4+9 \ln \left (t \right )\right )}{4374 t^{6}} \, dt}\\ &= - \int {\left (\frac {\cos \left (3 \ln \left (t \right )\right ) \left (-4+9 \ln \left (t \right )\right )}{243}\right ) \, dt}\\ &= -\frac {\left (\frac {8}{25}+\frac {9 \ln \left (t \right )}{10}\right ) t \cos \left (3 \ln \left (t \right )\right )}{243}+\frac {\left (\frac {87}{50}-\frac {27 \ln \left (t \right )}{10}\right ) t \sin \left (3 \ln \left (t \right )\right )}{243} \end{align*}

\begin{align*} U_2 &= (-1)^{4-2} \int { \frac {F(t) W_2(t) }{a W(t)} \, dt}\\ &= (-1)^{2} \int { \frac { \left (\cos \left (3 \ln \left (t \right )\right )\right ) \left (162 t^{6}\right )}{\left (t^{4}\right ) \left (4374 t^{2}\right )} \, dt} \\ &= \int { \frac {162 \cos \left (3 \ln \left (t \right )\right ) t^{6}}{4374 t^{6}} \, dt}\\ &= \int {\left (\frac {\cos \left (3 \ln \left (t \right )\right )}{27}\right ) \, dt}\\ &= \frac {\cos \left (3 \ln \left (t \right )\right ) t}{270}+\frac {t \sin \left (3 \ln \left (t \right )\right )}{90} \end{align*}

\begin{align*} U_3 &= (-1)^{4-3} \int { \frac {F(t) W_3(t) }{a W(t)} \, dt}\\ &= (-1)^{1} \int { \frac { \left (\cos \left (3 \ln \left (t \right )\right )\right ) \left (81 t^{3}\right )}{\left (t^{4}\right ) \left (4374 t^{2}\right )} \, dt} \\ &= - \int { \frac {81 \cos \left (3 \ln \left (t \right )\right ) t^{3}}{4374 t^{6}} \, dt}\\ &= - \int {\left (\frac {\cos \left (3 \ln \left (t \right )\right )}{54 t^{3}}\right ) \, dt}\\ &= -\frac {-\frac {1}{351}+\frac {\tan \left (\frac {3 \ln \left (t \right )}{2}\right )^{2}}{351}+\frac {\tan \left (\frac {3 \ln \left (t \right )}{2}\right )}{117}}{\left (1+\tan \left (\frac {3 \ln \left (t \right )}{2}\right )^{2}\right ) t^{2}} \end{align*}

\begin{align*} U_4 &= (-1)^{4-4} \int { \frac {F(t) W_4(t) }{a W(t)} \, dt}\\ &= (-1)^{0} \int { \frac { \left (\cos \left (3 \ln \left (t \right )\right )\right ) \left (\frac {9}{t^{3}}\right )}{\left (t^{4}\right ) \left (4374 t^{2}\right )} \, dt} \\ &= \int { \frac {\frac {9 \cos \left (3 \ln \left (t \right )\right )}{t^{3}}}{4374 t^{6}} \, dt}\\ &= \int {\left (\frac {\cos \left (3 \ln \left (t \right )\right )}{486 t^{9}}\right ) \, dt}\\ &= \frac {\left (-\frac {2}{17739}-\frac {i}{23652}\right ) t^{3 i}}{t^{8}}+\frac {\left (-\frac {2}{17739}+\frac {i}{23652}\right ) t^{-3 i}}{t^{8}} \end{align*}

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

\[ x_p = U_1 x_1+U_2 x_2+U_3 x_3+U_4 x_4 \]

Hence

\begin{equation*} \begin {split} x_p &= \left (-\frac {\left (\frac {8}{25}+\frac {9 \ln \left (t \right )}{10}\right ) t \cos \left (3 \ln \left (t \right )\right )}{243}+\frac {\left (\frac {87}{50}-\frac {27 \ln \left (t \right )}{10}\right ) t \sin \left (3 \ln \left (t \right )\right )}{243}\right ) \left (\frac {1}{t}\right ) \\ &+\left (\frac {\cos \left (3 \ln \left (t \right )\right ) t}{270}+\frac {t \sin \left (3 \ln \left (t \right )\right )}{90}\right ) \left (\frac {\ln \left (t \right )}{t}\right ) \\ &+\left (-\frac {-\frac {1}{351}+\frac {\tan \left (\frac {3 \ln \left (t \right )}{2}\right )^{2}}{351}+\frac {\tan \left (\frac {3 \ln \left (t \right )}{2}\right )}{117}}{\left (1+\tan \left (\frac {3 \ln \left (t \right )}{2}\right )^{2}\right ) t^{2}}\right ) \left (t^{2}\right ) \\ &+\left (\frac {\left (-\frac {2}{17739}-\frac {i}{23652}\right ) t^{3 i}}{t^{8}}+\frac {\left (-\frac {2}{17739}+\frac {i}{23652}\right ) t^{-3 i}}{t^{8}}\right ) \left (t^{8}\right ) \end {split} \end{equation*}

Therefore the particular solution is

\[ x_p = \left (\frac {31}{47450}-\frac {141 i}{94900}\right ) t^{-3 i} t^{6 i}+\left (\frac {31}{47450}+\frac {141 i}{94900}\right ) t^{-3 i} \]

Therefore the general solution is

\begin{align*} x &= x_h + x_p \\ &= \left (\frac {c_1}{t}+\frac {c_2 \ln \left (t \right )}{t}+c_3 \,t^{2}+c_4 \,t^{8}\right ) + \left (\left (\frac {31}{47450}-\frac {141 i}{94900}\right ) t^{-3 i} t^{6 i}+\left (\frac {31}{47450}+\frac {141 i}{94900}\right ) t^{-3 i}\right ) \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t^{4} x^{\prime \prime \prime \prime }-2 t^{3} x^{\prime \prime \prime }-20 t^{2} x^{\prime \prime }+12 x^{\prime } t +16 x=\cos \left (3 \ln \left (t \right )\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & x^{\prime \prime \prime \prime } \end {array} \]

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 
-> Calling odsolve with the ODE`, diff(diff(diff(_b(_a), _a), _a), _a) = (_C1-16*_b(_a)*_a+2*_a^2*(diff(_b(_a), _a))+6*_a^3*(diff(di 
   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 
   -> Calling odsolve with the ODE`, diff(diff(_g(_f), _f), _f) = _C2+8*_g(_f)/_f^2+6*(diff(_g(_f), _f))/_f-(1/3)*_C1/_f^3+(1/10)*(- 
      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 
         checking if the LODE is of Euler type 
         <- LODE of Euler type successful 
      <- solving first the homogeneous part of the ODE successful 
   <- high order exact_linear_nonhomogeneous successful 
<- high order exact_linear_nonhomogeneous successful`
                                                                                    
                                                                                    
 

Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 43

dsolve(t^4*diff(diff(diff(diff(x(t),t),t),t),t)-2*t^3*diff(diff(diff(x(t),t),t),t)-20*t^2*diff(diff(x(t),t),t)+12*diff(x(t),t)*t+16*x(t) = cos(3*ln(t)), 
       x(t),singsol=all)
 

\[ x = \frac {\left (15066+34263 i\right ) t^{1-3 i}+\left (15066-34263 i\right ) t^{1+3 i}[phone] t^{9} c_3 -1281150 c_2 \,t^{3}+854100 c_1 \ln \left (t \right )+94900 c_1 [phone] c_4}{23060700 t} \]

Mathematica DSolve solution

Solving time : 0.079 (sec)
Leaf size : 48

DSolve[{t^4*D[x[t],{t,4}]-2*t^3*D[x[t],{t,3}]-20*t^2*D[x[t],{t,2}]+12*t*D[x[t],t]+16*x[t]==Cos[3*Log[t]],{}}, 
       x[t],t,IncludeSingularSolutions->True]
 

\[ x(t)\to \frac {c_4 t^9+c_3 t^3+c_2 \log (t)+c_1}{t}+\frac {141 \sin (3 \log (t))}{47450}+\frac {31 \cos (3 \log (t))}{23725} \]