Internal problem ID [9878]
Internal file name [OUTPUT/8825_Monday_June_06_2022_05_34_32_AM_13997738/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 4, linear fourth order
Problem number: 1556.
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]]
\[ \boxed {x^{2} y^{\prime \prime \prime \prime }+8 x y^{\prime \prime \prime }+12 y^{\prime \prime }=0} \] 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 )+8 v^{\prime \prime }\left (x \right ) x +12 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 )+8 w^{\prime }\left (x \right ) x +12 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 \[ x^{2}(r(r-1))x^{r-2}+8 x r x^{r-1}+12 x^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}+8 r\,x^{r}+12 x^{r} = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )+8 r+12 = 0 \] Or \[ r^{2}+7 r +12 = 0 \tag {1} \] 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 &= -4\\ r_2 &= -3 \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 \[ w \left (x \right ) = \frac {c_{1}}{x^{4}}+\frac {c_{2}}{x^{3}} \] But since \(v^{\prime }\left (x \right ) = w \left (x \right )\) then we now need to solve the ode \(v^{\prime }\left (x \right ) = \frac {c_{1}}{x^{4}}+\frac {c_{2}}{x^{3}}\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{2} x +c_{1}}{x^{4}}\,\mathop {\mathrm {d}x}}\\ &= \frac {-\frac {c_{2} x}{2}-\frac {c_{1}}{3}}{x^{3}}+c_{3} \end {align*}
But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {-\frac {c_{2} x}{2}-\frac {c_{1}}{3}}{x^{3}}+c_{3}\). Integrating both sides gives \begin {align*} y &= \int { -\frac {-6 c_{3} x^{3}+3 c_{2} x +2 c_{1}}{6 x^{3}}\,\mathop {\mathrm {d}x}}\\ &= c_{3} x +\frac {c_{1}}{6 x^{2}}+\frac {c_{2}}{2 x}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x +\frac {c_{1}}{6 x^{2}}+\frac {c_{2}}{2 x}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = c_{3} x +\frac {c_{1}}{6 x^{2}}+\frac {c_{2}}{2 x}+c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime } x +12 y^{\prime \prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=-\frac {4 \left (2 y^{\prime \prime \prime } x +3 y^{\prime \prime }\right )}{x^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+\frac {8 y^{\prime \prime \prime }}{x}+\frac {12 y^{\prime \prime }}{x^{2}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime } x +12 y^{\prime \prime }=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+t^{\prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {3rd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{3}+3 t^{\prime }\left (x \right ) t^{\prime \prime }\left (x \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+t^{\prime \prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {4th}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\left (\frac {d^{4}}{d t^{4}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{4}+3 {t^{\prime }\left (x \right )}^{2} t^{\prime \prime }\left (x \right ) \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )+3 {t^{\prime \prime }\left (x \right )}^{2} \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+3 \left (t^{\prime \prime \prime }\left (x \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) t^{\prime }\left (x \right ) t^{\prime \prime }\left (x \right )\right ) t^{\prime }\left (x \right )+t^{\prime \prime \prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{\prime }\left (x \right ) t^{\prime \prime \prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\frac {\frac {d^{4}}{d t^{4}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}\right )}{x}-\frac {6 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{4}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {\frac {d^{4}}{d t^{4}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}\right )}{x}-\frac {6 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{4}}\right )+8 \left (\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}}\right ) x +\frac {12 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{2}}-\frac {12 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{2}}=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\frac {d^{4}}{d t^{4}}y \left (t \right )+2 \frac {d^{3}}{d t^{3}}y \left (t \right )-\frac {d^{2}}{d t^{2}}y \left (t \right )-2 \frac {d}{d t}y \left (t \right )}{x^{2}}=0 \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{4}}y \left (t \right )=-2 \frac {d^{3}}{d t^{3}}y \left (t \right )+\frac {d^{2}}{d t^{2}}y \left (t \right )+2 \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{4}}y \left (t \right )+2 \frac {d^{3}}{d t^{3}}y \left (t \right )-\frac {d^{2}}{d t^{2}}y \left (t \right )-2 \frac {d}{d t}y \left (t \right )=0 \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=\frac {d}{d t}y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=\frac {d^{2}}{d t^{2}}y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=\frac {d^{3}}{d t^{3}}y \left (t \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d t}y_{4}\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y_{4}\left (t \right )=-2 y_{4}\left (t \right )+y_{3}\left (t \right )+2 y_{2}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=\frac {d}{d t}y_{1}\left (t \right ), y_{3}\left (t \right )=\frac {d}{d t}y_{2}\left (t \right ), y_{4}\left (t \right )=\frac {d}{d t}y_{3}\left (t \right ), \frac {d}{d t}y_{4}\left (t \right )=-2 y_{4}\left (t \right )+y_{3}\left (t \right )+2 y_{2}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \\ y_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 1 & -2 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 2 & 1 & -2 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}={\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+c_{4} {\moverset {\rightarrow }{y}}_{4} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-t}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+c_{4} {\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{3} \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=-\frac {\left (-8 c_{4} {\mathrm e}^{3 t}-8 c_{3} {\mathrm e}^{2 t}+8 c_{2} {\mathrm e}^{t}+c_{1} \right ) {\mathrm e}^{-2 t}}{8} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=-\frac {-8 c_{4} x^{3}-8 c_{3} x^{2}+8 c_{2} x +c_{1}}{8 x^{2}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=c_{4} x +c_{3} -\frac {c_{2}}{x}-\frac {c_{1}}{8 x^{2}} \end {array} \]
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: 19
dsolve(x^2*diff(diff(diff(diff(y(x),x),x),x),x)+8*x*diff(diff(diff(y(x),x),x),x)+12*diff(diff(y(x),x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +\frac {c_{2}}{x^{2}}+\frac {c_{3}}{x}+c_{4} x \]
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 27
DSolve[12*y''[x] + 8*x*Derivative[3][y][x] + x^2*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {3 c_2 x+c_1}{6 x^2}+c_4 x+c_3 \]