Internal problem ID [6852]
Internal file name [OUTPUT/6099_Friday_July_29_2022_03_09_12_AM_9550685/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 35.
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]]
\[ \boxed {x^{2} y^{\prime \prime }-y^{\prime } \left (3 x -2 y^{\prime }\right )=0} \]
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 p \left (x \right )-3 x \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 u \left (x \right ) x -3 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 {2 u \left (u -1\right )}{x} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=u \left (u -1\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (u -1\right )} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{u \left (u -1\right )} \,du} &= \int {-\frac {2}{x} \,d x} \\ -\ln \left (u \right )+\ln \left (u -1\right )&=-2 \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}^{-2 \ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {u -1}{u} &= \frac {c_{3}}{x^{2}} \end {align*}
Therefore the solution \(p \left (x \right )\) is \begin {align*} p \left (x \right )&=x u\\ &=-\frac {x^{3}}{-x^{2}+c_{3}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {x^{3}}{-x^{2}+c_{3}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{3}}{-x^{2}+c_{3}}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+\frac {c_{3} \ln \left (x^{2}-c_{3} \right )}{2}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+\frac {c_{3} \ln \left (x^{2}-c_{3} \right )}{2}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+\frac {c_{3} \ln \left (x^{2}-c_{3} \right )}{2}+c_{4} \] 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)*(-3*_a+2*_b(_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: 0.062 (sec). Leaf size: 22
dsolve(x^2*diff(y(x),x$2)=diff(y(x),x)*(3*x-2*diff(y(x),x)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2}}{2}+\frac {c_{1} \ln \left (x^{2}-c_{1} \right )}{2}+c_{2} \]
✓ Solution by Mathematica
Time used: 0.333 (sec). Leaf size: 28
DSolve[x^2*y''[x]==y'[x]*(3*x-2*y'[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (x^2-c_1 \log \left (x^2+c_1\right )+2 c_2\right ) \]