Internal problem ID [9882]
Internal file name [OUTPUT/8829_Monday_June_06_2022_05_34_59_AM_15468329/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1560.
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_high_order, _missing_y]]
\[ \boxed {x^{3} y^{\prime \prime \prime \prime }+6 x^{2} y^{\prime \prime \prime }+6 x y^{\prime \prime }=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} x^{2} v^{\prime \prime \prime }\left (x \right )+6 x v^{\prime \prime }\left (x \right )+6 v^{\prime }\left (x \right ) = 0 \end {align*}
Since \(v \left (x \right )\) is missing from the ode then we can use the substitution \(v^{\prime }\left (x \right ) = w \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} x^{2} w^{\prime \prime }\left (x \right )+6 x w^{\prime }\left (x \right )+6 w \left (x \right ) = 0 \end {align*}
This is Euler second order ODE. Let the solution be \(w \left (x \right ) = x^r\), then \(w'=r x^{r-1}\) and \(w''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives \[ x^{2}(r(r-1))x^{r-2}+6 x r x^{r-1}+6 x^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}+6 r\,x^{r}+6 x^{r} = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )+6 r+6 = 0 \] Or \[ r^{2}+5 r +6 = 0 \tag {1} \] Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are \begin {align*} r_1 &= -3\\ r_2 &= -2 \end {align*}
Since the roots are real and distinct, then the general solution is \[ w \left (x \right )= c_{1} w_1 + c_{2} w_2 \] Where \(w_1 = x^{r_1}\) and \(w_2 = x^{r_2} \). Hence \[ w \left (x \right ) = \frac {c_{1}}{x^{3}}+\frac {c_{2}}{x^{2}} \] But since \(v^{\prime }\left (x \right ) = w \left (x \right )\) then we now need to solve the ode \(v^{\prime }\left (x \right ) = \frac {c_{1}}{x^{3}}+\frac {c_{2}}{x^{2}}\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{2} x +c_{1}}{x^{3}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {2 c_{2} x +c_{1}}{2 x^{2}}+c_{3} \end {align*}
But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = -\frac {2 c_{2} x +c_{1}}{2 x^{2}}+c_{3}\). Integrating both sides gives \begin {align*} y &= \int { -\frac {-2 c_{3} x^{2}+2 c_{2} x +c_{1}}{2 x^{2}}\,\mathop {\mathrm {d}x}}\\ &= c_{3} x -c_{2} \ln \left (x \right )+\frac {c_{1}}{2 x}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x -c_{2} \ln \left (x \right )+\frac {c_{1}}{2 x}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = c_{3} x -c_{2} \ln \left (x \right )+\frac {c_{1}}{2 x}+c_{4} \] 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: 18
dsolve(x^3*diff(diff(diff(diff(y(x),x),x),x),x)+6*x^2*diff(diff(diff(y(x),x),x),x)+6*x*diff(diff(y(x),x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +\ln \left (x \right ) c_{2} +\frac {c_{3}}{x}+c_{4} x \]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 27
DSolve[6*x*y''[x] + 6*x^2*Derivative[3][y][x] + x^3*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1}{2 x}+c_4 x-c_2 \log (x)+c_3 \]