Internal problem ID [15465]
Internal file name [OUTPUT/15466_Wednesday_May_08_2024_04_00_55_PM_99259692/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 17. Boundary value problems. Exercises page
163
Problem number: 722.
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]]
\[ \boxed {x^{2} y^{\prime \prime \prime \prime }+4 x y^{\prime \prime \prime }+2 y^{\prime \prime }=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = 0] \end {align*}
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 )+4 x v^{\prime \prime }\left (x \right )+2 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 )+4 x w^{\prime }\left (x \right )+2 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 \[ x^{2}(r(r-1))x^{r-2}+4 x r x^{r-1}+2 x^{r} = 0 \] Simplifying gives \[ r \left (r -1\right )x^{r}+4 r\,x^{r}+2 x^{r} = 0 \] Since \(x^{r}\neq 0\) then dividing throughout by \(x^{r}\) gives \[ r \left (r -1\right )+4 r+2 = 0 \] Or \[ r^{2}+3 r +2 = 0 \tag {1} \] 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 &= -2\\ r_2 &= -1 \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 \[ w \left (x \right ) = \frac {c_{1}}{x^{2}}+\frac {c_{2}}{x} \] But since \(v^{\prime }\left (x \right ) = w \left (x \right )\) then we now need to solve the ode \(v^{\prime }\left (x \right ) = \frac {c_{1}}{x^{2}}+\frac {c_{2}}{x}\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{2} x +c_{1}}{x^{2}}\,\mathop {\mathrm {d}x}}\\ &= c_{2} \ln \left (x \right )-\frac {c_{1}}{x}+c_{3} \end {align*}
But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = c_{2} \ln \left (x \right )-\frac {c_{1}}{x}+c_{3}\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{2} \ln \left (x \right ) x +c_{3} x -c_{1}}{x}\,\mathop {\mathrm {d}x}}\\ &= c_{2} \ln \left (x \right ) x -c_{1} \ln \left (x \right )-c_{2} x +c_{3} x +c_{4} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} \ln \left (x \right ) x -c_{1} \ln \left (x \right )-c_{2} x +c_{3} x +c_{4} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = -c_{2} +c_{3} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} \ln \left (x \right )-\frac {c_{1}}{x}+c_{3} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = -c_{1} +c_{3}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}\}\). Solving for the constants gives \begin {align*} c_{1}&=c_{3}\\ c_{2}&=c_{3} +c_{4} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = x \ln \left (x \right ) c_{3} +x \ln \left (x \right ) c_{4} -c_{3} \ln \left (x \right )-c_{4} x +c_{4} \end {align*}
Which simplifies to \[ y = \left (c_{4} x +c_{3} \left (x -1\right )\right ) \ln \left (x \right )-c_{4} \left (x -1\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{4} x +c_{3} \left (x -1\right )\right ) \ln \left (x \right )-c_{4} \left (x -1\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (c_{4} x +c_{3} \left (x -1\right )\right ) \ln \left (x \right )-c_{4} \left (x -1\right ) \] 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: 0.015 (sec). Leaf size: 22
dsolve([x^2*diff(y(x),x$4)+4*x*diff(y(x),x$3)+2*diff(y(x),x$2)=0,y(1) = 0, D(y)(1) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \left (-c_{3} +\left (x -1\right ) c_{4} \right ) \ln \left (x \right )+c_{3} \left (x -1\right ) \]
✓ Solution by Mathematica
Time used: 0.022 (sec). Leaf size: 29
DSolve[{x^2*y''''[x]+4*x*y'''[x]+2*y''[x]==0,{y[1]==0,y'[1]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (c_1-c_2) (x-1)+(c_2 x-c_1) \log (x) \]