2.3.18 Problem 18

Internal problem ID [8876]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 18
Date solved : Sunday, March 30, 2025 at 01:45:24 PM
CAS classification : [[_3rd_order, _with_linear_symmetries]]

Solved as higher order Euler type ode

Time used: 0.154 (sec)

The ode can be normalized and rewritten as Euler ode.

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

$$\begin{array}{rl}{y}^{\prime }& =\lambda x^{\lambda -1}\\ y^{\prime \prime }& =\lambda (\lambda -1)x^{\lambda -2}\\ y^{\prime \prime \prime }& =\lambda (\lambda -1)(\lambda -2)x^{\lambda -3}\end{array}$$

Substituting these back into

$$x^4{y}^{\prime \prime \prime }+x^3{y}^{\prime \prime }+x^2{y}^{\prime }+xy=0$$

gives

$$x\lambda x^{\lambda -1}+x^2\lambda (\lambda -1)x^{\lambda -2}+x^3\lambda (\lambda -1)(\lambda -2)x^{\lambda -3}+x^{\lambda }=0$$

Which simplifies to

$$\lambda x^{\lambda }+\lambda (\lambda -1)x^{\lambda }+\lambda (\lambda -1)(\lambda -2)x^{\lambda }+x^{\lambda }=0$$

And since $x^{\lambda }\ne 0$ then dividing through by $x^{\lambda }$, the above becomes

$$\lambda +\lambda (\lambda -1)+\lambda (\lambda -1)(\lambda -2)+1=0$$

Simplifying gives the characteristic equation as

$${\lambda }^3-2{\lambda }^2+2\lambda +1=0$$

Solving the above gives the following roots

$$\begin{array}{rl}{\lambda }_1& =-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}\\ {\lambda }_2& =\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}+\frac{i\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}\\ {\lambda }_3& =\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}-\frac{i\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}\end{array}$$

This table summarises the result

root	multiplicity	type of root
$\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}\pm -\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}i$	$1$	complex conjugate root
$-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}$	$1$	real root

The solution is generated by going over the above table. For each real root $\lambda$ of multiplicity one generates a $c_1{x}^{\lambda }$ basis solution. Each real root of multiplicty two, generates $c_1{x}^{\lambda }$ and $c_2{x}^{\lambda }\ln \left(x\right)$ basis solutions. Each real root of multiplicty three, generates $c_1{x}^{\lambda }$ and $c_2{x}^{\lambda }\ln \left(x\right)$ and $c_3{x}^{\lambda }\ln {\left(x\right)}^2$ basis solutions, and so on. Each complex root $\alpha \pm i\beta$ of multiplicity one generates $x^{\alpha }(c_1\cos (\beta \ln \left(x\right))+c_2\sin (\beta \ln \left(x\right)))$ basis solutions. And each complex root $\alpha \pm i\beta$ of multiplicity two generates $\ln \left(x\right)x^{\alpha }(c_1\cos (\beta \ln \left(x\right))+c_2\sin (\beta \ln \left(x\right)))$ basis solutions. And each complex root $\alpha \pm i\beta$ of multiplicity three generates $\ln {\left(x\right)}^2{x}^{\alpha }(c_1\cos (\beta \ln \left(x\right))+c_2\sin (\beta \ln \left(x\right)))$ basis solutions. And so on. Using the above show that the solution is

$$y=x^{\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}}(c_1\cos \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})\ln \left(x\right)}{2}\right)-c_2\sin \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})\ln \left(x\right)}{2}\right))+c_3{x}^{-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}}$$

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

$$\begin{array}{rl}{y}_1& =x^{\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}}\cos \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})\ln \left(x\right)}{2}\right)\\ y_2& =-x^{\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}}\sin \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})\ln \left(x\right)}{2}\right)\\ y_3& =x^{-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}}\end{array}$$

✓ Maple. Time used: 0.004 (sec). Leaf size: 184

ode:=x^4*diff(diff(diff(y(x),x),x),x)+x^3*diff(diff(y(x),x),x)+x^2*diff(y(x),x)+x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=c_1{x}^{-\frac{{(188+12\sqrt{249})}^{2/3}-4{(188+12\sqrt{249})}^{1/3}-8}{6{(188+12\sqrt{249})}^{1/3}}}+c_2{x}^{\frac{-8+{(188+12\sqrt{249})}^{2/3}+8{(188+12\sqrt{249})}^{1/3}}{12{(188+12\sqrt{249})}^{1/3}}}\sin \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)+c_3{x}^{\frac{-8+{(188+12\sqrt{249})}^{2/3}+8{(188+12\sqrt{249})}^{1/3}}{12{(188+12\sqrt{249})}^{1/3}}}\cos \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)$$

Maple trace

Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
<- LODE of Euler type successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & x^4\left(\frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)\right)+x^3\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)+x^2\left(\frac{d}{dx}y\left(x\right)\right)+xy\left(x\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}3\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Isolate 3rd derivative}\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)=-\frac{y\left(x\right)}{x^3}-\frac{\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)x+\frac{d}{dx}y\left(x\right)}{x^2}\\ \bullet & & \text{Group terms with}y\left(x\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)+\frac{\frac{d}{dx}\frac{d}{dx}y\left(x\right)}{x}+\frac{\frac{d}{dx}y\left(x\right)}{x^2}+\frac{y\left(x\right)}{x^3}=0\\ \bullet & & \text{Multiply by denominators of the ODE}\\ & & \left(\frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)\right)x^3+\left(\frac{d}{dx}\frac{d}{dx}y\left(x\right)\right)x^2+\left(\frac{d}{dx}y\left(x\right)\right)x+y\left(x\right)=0\\ \bullet & & \text{Make a change of variables}\\ & & t=\ln \left(x\right)\\ ◻& & \text{Substitute the change of variables back into the ODE}\\ & \circ & \text{Calculate the}\text{1st}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}y\left(t\right)\right)\left(\frac{d}{dx}t\left(x\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}y\left(t\right)}{x}\\ & \circ & \text{Calculate the}\text{2nd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right){\left(\frac{d}{dx}t\left(x\right)\right)}^2+\left(\frac{d}{dx}\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dt}y\left(t\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)=\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2}\\ & \circ & \text{Calculate the}\text{3rd}\text{derivative of}\text{y}\text{with respect to}\text{x}\text{, using the chain rule}\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)=\left(\frac{d}{dt}\frac{d^2}{d{t}^2}y\left(t\right)\right){\left(\frac{d}{dx}t\left(x\right)\right)}^3+3\left(\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dx}\frac{d}{dx}t\left(x\right)\right)\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+\left(\frac{d}{dx}\frac{d^2}{d{x}^2}t\left(x\right)\right)\left(\frac{d}{dt}y\left(t\right)\right)\\ & \circ & \text{Compute derivative}\\ & & \frac{d}{dx}\frac{d^2}{d{x}^2}y\left(x\right)=\frac{\frac{d}{dt}\frac{d^2}{d{t}^2}y\left(t\right)}{x^3}-\frac{3\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)}{x^3}+\frac{2\left(\frac{d}{dt}y\left(t\right)\right)}{x^3}\\ & & \text{Substitute the change of variables back into the ODE}\\ & & (\frac{\frac{d}{dt}\frac{d^2}{d{t}^2}y\left(t\right)}{x^3}-\frac{3\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)}{x^3}+\frac{2\left(\frac{d}{dt}y\left(t\right)\right)}{x^3})x^3+(\frac{\frac{d}{dt}\frac{d}{dt}y\left(t\right)}{x^2}-\frac{\frac{d}{dt}y\left(t\right)}{x^2})x^2+\frac{d}{dt}y\left(t\right)+y\left(t\right)=0\\ \bullet & & \text{Simplify}\\ & & \frac{d}{dt}\frac{d^2}{d{t}^2}y\left(t\right)-2\frac{d}{dt}\frac{d}{dt}y\left(t\right)+2\frac{d}{dt}y\left(t\right)+y\left(t\right)=0\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^3-2r^2+2r+1=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=[-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3},\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}+\frac{\mathrm{I}\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2},\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}-\frac{\mathrm{I}\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}]\\ \bullet & & \text{Solution from}r=-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}\\ & & y_1\left(t\right)={\mathrm{e}}^{(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}+\frac{4}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3})t}\\ \bullet & & \text{Solutions from}r=\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}+\frac{\mathrm{I}\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}\text{and}r=\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3}-\frac{\mathrm{I}\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})}{2}\\ & & [y_2\left(t\right)=-{\mathrm{e}}^{(\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3})t}\sin \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})t}{2}\right),y_3\left(t\right)={\mathrm{e}}^{(\frac{{(188+12\sqrt{249})}^{1/3}}{12}-\frac{2}{3{(188+12\sqrt{249})}^{1/3}}+\frac{2}{3})t}\cos \left(\frac{\sqrt{3}(-\frac{{(188+12\sqrt{249})}^{1/3}}{6}-\frac{4}{3{(188+12\sqrt{249})}^{1/3}})t}{2}\right)]\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{y}_1\left(t\right)+\mathit{C}\mathit{2}{y}_2\left(t\right)+\mathit{C}\mathit{3}{y}_3\left(t\right)\\ \bullet & & \text{Substitute in solutions and simplify}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{({(188+12\sqrt{249})}^{2/3}-4{(188+12\sqrt{249})}^{1/3}-8)t}{6{(188+12\sqrt{249})}^{1/3}}}+{\mathrm{e}}^{\frac{({(188+12\sqrt{249})}^{2/3}+8{(188+12\sqrt{249})}^{1/3}-8)t}{12{(188+12\sqrt{249})}^{1/3}}}(\sin \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)t}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)\mathit{C}\mathit{2}+\cos \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)t}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)\mathit{C}\mathit{3})\\ \bullet & & \text{Change variables back using}t=\ln \left(x\right)\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{({(188+12\sqrt{249})}^{2/3}-4{(188+12\sqrt{249})}^{1/3}-8)\ln \left(x\right)}{6{(188+12\sqrt{249})}^{1/3}}}+{\mathrm{e}}^{\frac{({(188+12\sqrt{249})}^{2/3}+8{(188+12\sqrt{249})}^{1/3}-8)\ln \left(x\right)}{12{(188+12\sqrt{249})}^{1/3}}}(\sin \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)\mathit{C}\mathit{2}+\cos \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)\mathit{C}\mathit{3})\\ \bullet & & \text{Simplify}\\ & & y\left(x\right)=x^{2/3}(\mathit{C}\mathit{3}{x}^{\frac{{(188+12\sqrt{249})}^{2/3}-8}{12{(188+12\sqrt{249})}^{1/3}}}\cos \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)+\mathit{C}\mathit{2}{x}^{\frac{{(188+12\sqrt{249})}^{2/3}-8}{12{(188+12\sqrt{249})}^{1/3}}}\sin \left(\frac{\sqrt{3}({(188+12\sqrt{3}\sqrt{83})}^{2/3}+8)\ln \left(x\right)}{12{(188+12\sqrt{3}\sqrt{83})}^{1/3}}\right)+\mathit{C}\mathit{1}{x}^{-\frac{{(188+12\sqrt{249})}^{2/3}-8}{6{(188+12\sqrt{249})}^{1/3}}})\end{array}$$

✓ Mathematica. Time used: 0.006 (sec). Leaf size: 81

ode=x^4*D[y[x],{x,3}]+x^3*D[y[x],{x,2}]+x^2*D[y[x],x]+x*y[x]== 0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1{x}^{\text{Root}[{\mathrm{\#}\text{1}}^3-2{\mathrm{\#}\text{1}}^2+2\mathrm{\#}\text{1}+1\mathrm{\&},1]}+c_3{x}^{\text{Root}[{\mathrm{\#}\text{1}}^3-2{\mathrm{\#}\text{1}}^2+2\mathrm{\#}\text{1}+1\mathrm{\&},3]}+c_2{x}^{\text{Root}[{\mathrm{\#}\text{1}}^3-2{\mathrm{\#}\text{1}}^2+2\mathrm{\#}\text{1}+1\mathrm{\&},2]}$$

✓ Sympy. Time used: 0.324 (sec). Leaf size: 54

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

$$y\left(x\right)=\frac{C_1}{x^{-\mathrm{CRootOf}(x^3-2x^2+2x+1,0)}}+C_2{x}^{\mathrm{CRootOf}(x^3-2x^2+2x+1,1)}+C_3{x}^{\mathrm{CRootOf}(x^3-2x^2+2x+1,2)}$$