2.7.11 Problem 14

Internal problem ID [19760]
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 : 14
Date solved : Friday, November 28, 2025 at 06:51:11 PM
CAS classification : [[_high_order, _missing_y]]

Solve

\begin{align*} x^{4} y^{\prime \prime \prime \prime }+x^{3} y^{\prime \prime \prime }-20 x^{2} y^{\prime \prime }+20 y^{\prime } x&=17 x^{6} \\ \end{align*}

Solved as higher order Euler type ode

Time used: 0.520 (sec)

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

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

Substituting these back into

\[ x^{4} y^{\prime \prime \prime \prime }+x^{3} y^{\prime \prime \prime }-20 x^{2} y^{\prime \prime }+20 y^{\prime } x = 17 x^{6} \]

gives

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

Which simplifies to

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

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

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

Simplifying gives the characteristic equation as

\[ \lambda ^{4}-5 \lambda ^{3}-12 \lambda ^{2}+36 \lambda = 0 \]

Solving the above gives the following roots

\begin{align*} \lambda _1 &= 0\\ \lambda _2 &= 2\\ \lambda _3 &= 6\\ \lambda _4 &= -3 \end{align*}

This table summarises the result

root	multiplicity	type of root
\(0\)	\(1\)	real root
\(2\)	\(1\)	real root
\(-3\)	\(1\)	real root
\(6\)	\(1\)	real root

The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1x^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) and \(c_3x^{\lambda } \ln \left (x \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(x^{\alpha } \left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (x \right ) x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (x \right )^{2} x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And so on. Using the above show that the solution is

\[ y = c_2 \,x^{2}+c_1 +\frac {c_3}{x^{3}}+c_4 \,x^{6} \]

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

\begin{align*} y_1 &= 1 \\ y_2 &= x^{2} \\ y_3 &= \frac {1}{x^{3}} \\ y_4 &= x^{6} \\ \end{align*}

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

\[ y = y_h + y_p \]

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

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

Now the particular solution to the given ODE is found

\[ x^{4} y^{\prime \prime \prime \prime }+x^{3} y^{\prime \prime \prime }-20 x^{2} y^{\prime \prime }+20 y^{\prime } x = 17 x^{6} \]

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^{2} & \frac {1}{x^{3}} & x^{6} \\ 0 & 2 x & -\frac {3}{x^{4}} & 6 x^{5} \\ 0 & 2 & \frac {12}{x^{5}} & 30 x^{4} \\ 0 & 0 & -\frac {60}{x^{6}} & 120 x^{3} \end {array}\right ] \\ |W| &= \frac {6480}{x} \end{align*}

The determinant simplifies to

\begin{align*} |W| &= \frac {6480}{x} \end{align*}

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

\begin{align*} W_1(x) &= \det \,\left [\begin {array}{ccc} x^{2} & \frac {1}{x^{3}} & x^{6} \\ 2 x & -\frac {3}{x^{4}} & 6 x^{5} \\ 2 & \frac {12}{x^{5}} & 30 x^{4} \end {array}\right ] \\ &= -180 x^{2} \end{align*}

\begin{align*} W_2(x) &= \det \,\left [\begin {array}{ccc} 1 & \frac {1}{x^{3}} & x^{6} \\ 0 & -\frac {3}{x^{4}} & 6 x^{5} \\ 0 & \frac {12}{x^{5}} & 30 x^{4} \end {array}\right ] \\ &= -162 \end{align*}

\begin{align*} W_3(x) &= \det \,\left [\begin {array}{ccc} 1 & x^{2} & x^{6} \\ 0 & 2 x & 6 x^{5} \\ 0 & 2 & 30 x^{4} \end {array}\right ] \\ &= 48 x^{5} \end{align*}

\begin{align*} W_4(x) &= \det \,\left [\begin {array}{ccc} 1 & x^{2} & \frac {1}{x^{3}} \\ 0 & 2 x & -\frac {3}{x^{4}} \\ 0 & 2 & \frac {12}{x^{5}} \end {array}\right ] \\ &= \frac {30}{x^{4}} \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 (17 x^{6}\right ) \left (-180 x^{2}\right )}{\left (x^{4}\right ) \left (\frac {6480}{x}\right )} \, dx} \\ &= - \int { \frac {-3060 x^{8}}{6480 x^{3}} \, dx}\\ &= - \int {\left (-\frac {17 x^{5}}{36}\right ) \, dx}\\ &= \frac {17 x^{6}}{216} \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 (17 x^{6}\right ) \left (-162\right )}{\left (x^{4}\right ) \left (\frac {6480}{x}\right )} \, dx} \\ &= \int { \frac {-2754 x^{6}}{6480 x^{3}} \, dx}\\ &= \int {\left (-\frac {17 x^{3}}{40}\right ) \, dx}\\ &= -\frac {17 x^{4}}{160} \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 (17 x^{6}\right ) \left (48 x^{5}\right )}{\left (x^{4}\right ) \left (\frac {6480}{x}\right )} \, dx} \\ &= - \int { \frac {816 x^{11}}{6480 x^{3}} \, dx}\\ &= - \int {\left (\frac {17 x^{8}}{135}\right ) \, dx}\\ &= -\frac {17 x^{9}}{1215} \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 (17 x^{6}\right ) \left (\frac {30}{x^{4}}\right )}{\left (x^{4}\right ) \left (\frac {6480}{x}\right )} \, dx} \\ &= \int { \frac {510 x^{2}}{6480 x^{3}} \, dx}\\ &= \int {\left (\frac {17}{216 x}\right ) \, dx}\\ &= \frac {17 \ln \left (x \right )}{216} \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 {17 x^{6}}{216}\right ) \\ &+\left (-\frac {17 x^{4}}{160}\right ) \left (x^{2}\right ) \\ &+\left (-\frac {17 x^{9}}{1215}\right ) \left (\frac {1}{x^{3}}\right ) \\ &+\left (\frac {17 \ln \left (x \right )}{216}\right ) \left (x^{6}\right ) \end {split} \end{equation*}

Therefore the particular solution is

\[ y_p = \frac {17 x^{6} \left (-19+36 \ln \left (x \right )\right )}{7776} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_2 \,x^{2}+c_1 +\frac {c_3}{x^{3}}+c_4 \,x^{6}\right ) + \left (\frac {17 x^{6} \left (-19+36 \ln \left (x \right )\right )}{7776}\right ) \\ \end{align*}

✓ Maple. Time used: 0.003 (sec). Leaf size: 41

ode:=x^4*diff(diff(diff(diff(y(x),x),x),x),x)+x^3*diff(diff(diff(y(x),x),x),x)-20*x^2*diff(diff(y(x),x),x)+20*diff(y(x),x)*x = 17*x^6; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {612 \ln \left (x \right ) x^{9}+\left (1296 c_3 -323\right ) x^{9}+3888 c_1 \,x^{5}+7776 c_4 \,x^{3}-2592 c_2}{7776 x^{3}} \]

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] 
-> Calling odsolve with the ODE, diff(diff(diff(_b(_a),_a),_a),_a) = -(-17*_a^5 
+diff(diff(_b(_a),_a),_a)*_a^2-20*diff(_b(_a),_a)*_a+20*_b(_a))/_a^3, _b(_a) 
   *** Sublevel 2 *** 
   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 is of Euler type 
   <- LODE of Euler type successful 
   Euler equation successful 
<- differential order: 4; linear nonhomogeneous with symmetry [0,1] successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d^{4}}{d x^{4}}y \left (x \right )\right )+x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )-20 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+20 x \left (\frac {d}{d x}y \left (x \right )\right )=17 x^{6} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right ) \end {array} \]

✓ Mathematica. Time used: 0.01 (sec). Leaf size: 49

ode=x^4*D[y[x],{x,4}]+x^3*D[y[x],{x,3}]-20*x^2*D[y[x],{x,2}]+20*x*D[y[x],x]==17*x^6; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
                                                                                         
                                                                                         
 

\begin{align*} y(x)&\to \frac {17}{216} x^6 \log (x)+\left (-\frac {323}{7776}+\frac {c_3}{6}\right ) x^6-\frac {c_1}{3 x^3}+\frac {c_2 x^2}{2}+c_4 \end{align*}

✓ Sympy. Time used: 0.222 (sec). Leaf size: 29

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-17*x**6 + x**4*Derivative(y(x), (x, 4)) + x**3*Derivative(y(x), (x, 3)) - 20*x**2*Derivative(y(x), (x, 2)) + 20*x*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = C_{1} + \frac {C_{2}}{x^{3}} + C_{3} x^{2} + C_{4} x^{6} + \frac {17 x^{6} \log {\left (x \right )}}{216} \]