Internal problem ID [6240]
Internal file name [OUTPUT/5488_Sunday_June_05_2022_03_41_15_PM_99919298/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.9. Reduction of Order. Page
38
Problem number: 1(d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x^{2} y^{\prime \prime }-2 y^{\prime } x -{y^{\prime }}^{2}=0} \]
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^{2} p^{\prime }\left (x \right )+\left (-2 x -p \left (x \right )\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Using the change of variables \(p \left (x \right ) = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} x^{2} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+\left (-2 x -u \left (x \right ) x \right ) u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (u +1\right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u \left (u +1\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (u +1\right )} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u \left (u +1\right )} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (u \right )-\ln \left (u +1\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u \right )-\ln \left (u +1\right )} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {u}{u +1} &= c_{3} x \end {align*}
Therefore the solution \(p \left (x \right )\) is \begin {align*} p \left (x \right )&=x u\\ &=-\frac {x^{2} c_{3}}{c_{3} x -1} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {x^{2} c_{3}}{c_{3} x -1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2} c_{3}}{c_{3} x -1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {x^{2}}{2}-\frac {x}{c_{3}}-\frac {\ln \left (c_{3} x -1\right )}{c_{3}^{2}}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{2}}{2}-\frac {x}{c_{3}}-\frac {\ln \left (c_{3} x -1\right )}{c_{3}^{2}}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{2}}{2}-\frac {x}{c_{3}}-\frac {\ln \left (c_{3} x -1\right )}{c_{3}^{2}}+c_{4} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = _b(_a)*(_b(_a)+2*_a)/_a^2, _b(_a), HINT = [[_a, _b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, _b]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 26
dsolve(x^2*diff(y(x),x$2)=2*x*diff(y(x),x)+(diff(y(x),x))^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{2}}{2}-c_{1} x -c_{1}^{2} \ln \left (x -c_{1} \right )+c_{2} \]
✓ Solution by Mathematica
Time used: 0.435 (sec). Leaf size: 41
DSolve[x^2*y''[x]==2*x*y'[x]+(y'[x])^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x^2}{2}-c_1 x-c_1{}^2 \log (x-c_1)+\frac {3 c_1{}^2}{2}+c_2 \]