Internal problem ID [15193]
Internal file name [OUTPUT/15194_Tuesday_April_23_2024_04_53_52_PM_3577603/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 14. Differential equations admitting of
depression of their order. Exercises page 107
Problem number: 336.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y", "second_order_ode_non_constant_coeff_transformation_on_B"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x \ln \left (x \right ) y^{\prime \prime }-y^{\prime }=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} x \ln \left (x \right ) p^{\prime }\left (x \right )-p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p}{x \ln \left (x \right )} \end {align*}
Where \(f(x)=\frac {1}{x \ln \left (x \right )}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {1}{x \ln \left (x \right )} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {\frac {1}{x \ln \left (x \right )} \,d x}\\ \ln \left (p \right )&=\ln \left (\ln \left (x \right )\right )+c_{1}\\ p&={\mathrm e}^{\ln \left (\ln \left (x \right )\right )+c_{1}}\\ &=c_{1} \ln \left (x \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{1} \ln \left (x \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{1} \ln \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= x c_{1} \ln \left (x \right )-c_{1} x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x c_{1} \ln \left (x \right )-c_{1} x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = x c_{1} \ln \left (x \right )-c_{1} x +c_{2} \] Verified OK.
Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}
This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}
This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}
And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}
If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).
This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= \ln \left (x \right ) x\\ B &= -1\\ C &= 0\\ F &= 0 \end {align*}
The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (\ln \left (x \right ) x\right ) \left (0\right ) + \left (-1\right ) \left (0\right ) + \left (0\right ) \left (-1\right ) \\ &=0 \end {align*}
Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} -\ln \left (x \right ) x v'' +\left ( 1\right ) v' & =0 \end {align*}
Now by applying \(v'=u\) the above becomes \begin {align*} -\ln \left (x \right ) x u^{\prime }\left (x \right )+u \left (x \right ) = 0 \end {align*}
Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u}{x \ln \left (x \right )} \end {align*}
Where \(f(x)=\frac {1}{x \ln \left (x \right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {1}{x \ln \left (x \right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {1}{x \ln \left (x \right )} \,d x}\\ \ln \left (u \right )&=\ln \left (\ln \left (x \right )\right )+c_{1}\\ u&={\mathrm e}^{\ln \left (\ln \left (x \right )\right )+c_{1}}\\ &=c_{1} \ln \left (x \right ) \end {align*}
The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=c_{1} \ln \left (x \right ) \end {align*}
Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { c_{1} \ln \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= x c_{1} \ln \left (x \right )-c_{1} x +c_{2} \end {align*}
Therefore the solution is \begin {align*} y(x) &= B v\\ &= \left (-1\right ) \left (x c_{1} \ln \left (x \right )-c_{1} x +c_{2}\right ) \\ &= -x c_{1} \ln \left (x \right )+c_{1} x -c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x c_{1} \ln \left (x \right )+c_{1} x -c_{2} \\ \end{align*}
Verification of solutions
\[ y = -x c_{1} \ln \left (x \right )+c_{1} x -c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \ln \left (x \right ) y^{\prime \prime }-y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & x \ln \left (x \right ) u^{\prime }\left (x \right )-u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {u \left (x \right )}{x \ln \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )}=\frac {1}{x \ln \left (x \right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )}d x =\int \frac {1}{x \ln \left (x \right )}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (x \right )\right )=\ln \left (\ln \left (x \right )\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{c_{1}} \ln \left (x \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{c_{1}} \ln \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \ln \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int {\mathrm e}^{c_{1}} \ln \left (x \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{c_{1}} \left (\ln \left (x \right ) x -x \right )+c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y <- LODE missing y successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 15
dsolve(x*ln(x)*diff(y(x),x$2)=diff(y(x),x),y(x), singsol=all)
\[ y = \ln \left (x \right ) c_{2} x -c_{2} x +c_{1} \]
✓ Solution by Mathematica
Time used: 0.046 (sec). Leaf size: 19
DSolve[x*Log[x]*y''[x]==y'[x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 (-x)+c_1 x \log (x)+c_2 \]