Internal problem ID [13672]
Internal file name [OUTPUT/12844_Saturday_February_17_2024_08_45_22_AM_61754550/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 20. Euler equations. Additional exercises page 382
Problem number: 20.4 (g).
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coefficients_of_type_Euler"
Maple gives the following as the ode type
[[_high_order, _with_linear_symmetries]]
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+2 x^{3} y^{\prime \prime \prime }+x^{2} y^{\prime \prime }-y^{\prime } x +y=0} \] 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 \[ x^{4} y^{\prime \prime \prime \prime }+2 x^{3} y^{\prime \prime \prime }+x^{2} y^{\prime \prime }-y^{\prime } x +y = 0 \] gives \[ -x \lambda \,x^{\lambda -1}+x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+2 x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4}+x^{\lambda } = 0 \] Which simplifies to \[ -\lambda \,x^{\lambda }+\lambda \left (\lambda -1\right ) x^{\lambda }+2 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda }+x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ -\lambda +\lambda \left (\lambda -1\right )+2 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )+1 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda -1\right )^{4} = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \(1\) | \(4\) | 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
\[ y = c_{1} x +c_{2} x \ln \left (x \right )+c_{3} \ln \left (x \right )^{2} x +c_{4} \ln \left (x \right )^{3} x \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= x\\ y_2 &= x \ln \left (x \right )\\ y_3 &= \ln \left (x \right )^{2} x\\ y_4 &= \ln \left (x \right )^{3} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{2} x \ln \left (x \right )+c_{3} \ln \left (x \right )^{2} x +c_{4} \ln \left (x \right )^{3} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{2} x \ln \left (x \right )+c_{3} \ln \left (x \right )^{2} x +c_{4} \ln \left (x \right )^{3} x \] Verified OK.
Maple trace
`Methods for high 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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 24
dsolve(x^4*diff(y(x),x$4)+2*x^3*diff(y(x),x$3)+x^2*diff(y(x),x$2)-x*diff(y(x),x)+y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x \left (c_{1} +c_{2} \ln \left (x \right )+c_{3} \ln \left (x \right )^{2}+c_{4} \ln \left (x \right )^{3}\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 29
DSolve[x^4*y''''[x]+2*x^3*y'''[x]+x^2*y''[x]-x*y'[x]+y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x \left (c_4 \log ^3(x)+c_3 \log ^2(x)+c_2 \log (x)+c_1\right ) \]