Internal problem ID [6823]
Internal file name [OUTPUT/6070_Thursday_July_28_2022_04_29_06_AM_15447379/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: 3.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+{y^{\prime }}^{2}-2 y^{\prime } x=0} \] With initial conditions \begin {align*} [y \left (2\right ) = 5, y^{\prime }\left (2\right ) = 2] \end {align*}
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 (p \left (x \right )-2 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 (u \left (x \right ) x -2 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 -1}{u} &= \frac {c_{3}}{x} \end {align*}
Therefore the solution \(p \left (x \right )\) is \begin {align*} p \left (x \right )&=u x\\ &=-\frac {x^{2}}{c_{3} -x} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=2\) and \(p=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = -\frac {4}{c_{3} -2} \end {align*}
The solutions are \begin {align*} c_{3} = 0 \end {align*}
Trying the constant \begin {align*} c_{3} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (x \right )&=x \end {align*}
The constant \(c_{3} = 0\) gives valid solution.
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+c_{4} \end {align*}
Initial conditions are used to solve for \(c_{4}\). Substituting \(x=2\) and \(y=5\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 5 = 2+c_{4} \end {align*}
The solutions are \begin {align*} c_{4} = 3 \end {align*}
Trying the constant \begin {align*} c_{4} = 3 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {x^{2}}{2}+3 \end {align*}
The constant \(c_{4} = 3\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+3 \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+3 \] 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: 0.015 (sec). Leaf size: 11
dsolve([x^2*diff(y(x),x$2)+diff(y(x),x)^2-2*x*diff(y(x),x)=0,y(2) = 5, D(y)(2) = 2],y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2}}{2}+3 \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 14
DSolve[{x^2*y''[x]+(y'[x])^2-2*x*y'[x]==0,{y[2]==5,y'[2]==2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (x^2+6\right ) \]