Internal problem ID [6856]
Internal file name [OUTPUT/6103_Friday_July_29_2022_03_09_22_AM_41838316/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: 39.
ODE order: 2.
ODE degree: 2.
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]]
\[ \boxed {{y^{\prime \prime }}^{2}-2 y^{\prime \prime }+{y^{\prime }}^{2}-2 y^{\prime } x=-x^{2}} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {1}{2}}, y^{\prime }\left (0\right ) = 1\right ] \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*} \left (p^{\prime }\left (x \right )-2\right ) p^{\prime }\left (x \right )+\left (p \left (x \right )-2 x \right ) p \left (x \right )+x^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Let \(p=p^{\prime }\left (x \right )\) the ode becomes \begin {align*} \left (p -2\right ) p +\left (p -2 x \right ) p = -x^{2} \end {align*}
Solving for \(p \left (x \right )\) from the above results in \begin {align*} p \left (x \right ) &= x +\sqrt {-p^{2}+2 p}\tag {1A}\\ p \left (x \right ) &= x -\sqrt {-p^{2}+2 p}\tag {2A} \end {align*}
This has the form \begin {align*} p=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=p'(x)\). Each of the above ode’s is dAlembert ode which is now solved. Solving ode 1A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(p \left (x \right )=x f + g\) to (1A) shows that \begin {align*} f &= 1\\ g &= \sqrt {-\left (p -2\right ) p} \end {align*}
Hence (2) becomes \begin {align*} p -1 = \frac {\left (-2 p +2\right ) p^{\prime }\left (x \right )}{2 \sqrt {-\left (p -2\right ) p}}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -1 = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=1 \end {align*}
Substituting these in (1A) gives \begin {align*} p \left (x \right )&=1+x \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {2 \left (p \left (x \right )-1\right ) \sqrt {-\left (p \left (x \right )-2\right ) p \left (x \right )}}{-2 p \left (x \right )+2}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\). Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {-\left (p -2\right ) p}}d p &= x +c_{1}\\ -\arcsin \left (p -1\right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=1-\sin \left (x +c_{1} \right ) \end {align*}
Substituing the above solution for \(p\) in (2A) gives \begin {align*} p \left (x \right ) = x +\sqrt {-\left (-1-\sin \left (x +c_{1} \right )\right ) \left (1-\sin \left (x +c_{1} \right )\right )}\\ \end {align*}
Solving ode 2A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(p \left (x \right )=x f + g\) to (1A) shows that \begin {align*} f &= 1\\ g &= -\sqrt {-\left (p -2\right ) p} \end {align*}
Hence (2) becomes \begin {align*} p -1 = -\frac {\left (-2 p +2\right ) p^{\prime }\left (x \right )}{2 \sqrt {-\left (p -2\right ) p}}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -1 = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=1 \end {align*}
Substituting these in (1A) gives \begin {align*} p \left (x \right )&=x -1 \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = -\frac {2 \left (p \left (x \right )-1\right ) \sqrt {-\left (p \left (x \right )-2\right ) p \left (x \right )}}{-2 p \left (x \right )+2}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\). Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {-\left (p -2\right ) p}}d p &= x +c_{2}\\ \arcsin \left (p -1\right )&=x +c_{2} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=1+\sin \left (x +c_{2} \right ) \end {align*}
Substituing the above solution for \(p\) in (2A) gives \begin {align*} p \left (x \right ) = x -\sqrt {-\left (-1+\sin \left (x +c_{2} \right )\right ) \left (1+\sin \left (x +c_{2} \right )\right )}\\ \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = -\sqrt {-\left (-1+\sin \left (c_{2} \right )\right ) \left (1+\sin \left (c_{2} \right )\right )} \end {align*}
The solutions are \begin {align*} c_{2} = \pi \end {align*}
Trying the constant \begin {align*} c_{2} = \pi \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (x \right )&=x -\sqrt {-\left (1+\sin \left (x \right )\right ) \left (\sin \left (x \right )-1\right )} \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{2} = \pi \) does not give valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = 1+x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { 1+x\,\mathop {\mathrm {d}x}}\\ &= \frac {x \left (x +2\right )}{2}+c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{2}} = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = {\frac {1}{2}} \end {align*}
Trying the constant \begin {align*} c_{3} = {\frac {1}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {1}{2} x^{2}+x +\frac {1}{2} \end {align*}
The constant \(c_{3} = {\frac {1}{2}}\) gives valid solution.
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x +\sqrt {-\left (-1-\sin \left (x +c_{1} \right )\right ) \left (1-\sin \left (x +c_{1} \right )\right )} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x +\sqrt {-\left (-1-\sin \left (x +c_{1} \right )\right ) \left (1-\sin \left (x +c_{1} \right )\right )}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}-\frac {2 \left (-1+\sin \left (x +c_{1} \right )\right )^{2} \left (1+\sin \left (x +c_{1} \right )\right )}{3 \cos \left (x +c_{1} \right ) \sqrt {-\left (1+\sin \left (x +c_{1} \right )\right ) \left (-1+\sin \left (x +c_{1} \right )\right )}}+c_{4} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{2}} = -\frac {2 \left (-1+\sin \left (c_{1} \right )\right )^{2} \left (1+\sin \left (c_{1} \right )\right )}{3 \cos \left (c_{1} \right ) \sqrt {-\left (1+\sin \left (c_{1} \right )\right ) \left (-1+\sin \left (c_{1} \right )\right )}}+c_{4} \end {align*}
The solutions are \begin {align*} c_{1} = -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right ) \end {align*}
Trying the constant \begin {align*} c_{1} = -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right ) \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {x^{2}}{2}-\frac {2 \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )^{2} \left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}{3 \cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right ) \sqrt {-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}}+c_{4} \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{1} = -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\) does not give valid solution.
\begin {align*} {\frac {1}{2}} = {\frac {1}{2}} \end {align*}
Which is valid for any constant of integration. Therefore keeping the constant in place. Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x -1 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x -1\,\mathop {\mathrm {d}x}}\\ &= \frac {x \left (x -2\right )}{2}+c_{5} \end {align*}
Initial conditions are used to solve for \(c_{5}\). Substituting \(x=0\) and \(y={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{2}} = c_{5} \end {align*}
The solutions are \begin {align*} c_{5} = {\frac {1}{2}} \end {align*}
Trying the constant \begin {align*} c_{5} = {\frac {1}{2}} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {1}{2} x^{2}-x +\frac {1}{2} \end {align*}
The constant \(c_{5} = {\frac {1}{2}}\) gives valid solution.
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x -\sqrt {-\left (1+\sin \left (x \right )\right ) \left (\sin \left (x \right )-1\right )} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x -\sqrt {-\left (1+\sin \left (x \right )\right ) \left (\sin \left (x \right )-1\right )}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+\frac {2 \left (\sin \left (x \right )-1\right )^{2} \left (1+\sin \left (x \right )\right )}{3 \cos \left (x \right ) \sqrt {-\left (1+\sin \left (x \right )\right ) \left (\sin \left (x \right )-1\right )}}+c_{6} \end {align*}
Initial conditions are used to solve for \(c_{6}\). Substituting \(x=0\) and \(y={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {1}{2}} = c_{6} +\frac {2}{3} \end {align*}
The solutions are \begin {align*} c_{6} = -{\frac {1}{6}} \end {align*}
Trying the constant \begin {align*} c_{6} = -{\frac {1}{6}} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {2 \sin \left (x \right )}{3}+\frac {x^{2}}{2}+\frac {1}{2} \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{6} = -{\frac {1}{6}}\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place. Initial conditions are used to solve for the constants of integration.
Looking at the Second solution \begin {align*} y = \frac {x^{2}}{2}-\frac {2 \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )^{2} \left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}{3 \cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right ) \sqrt {-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}}+c_{4} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = {\frac {1}{2}}\) and \(x = 0\) in the above gives \begin {align*} {\frac {1}{2}} = {\frac {1}{2}}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = x -\frac {4 \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}{3 \sqrt {-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}}-\frac {2 \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )^{2}}{3 \sqrt {-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}}-\frac {2 \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )^{2} \left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )}{3 \cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )^{2} \sqrt {-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )}}+\frac {\left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )^{2} \left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )-\cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )\right )}{3 \cos \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right ) \left (-\left (1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right ) \left (-1+\sin \left (x -\arcsin \left (-\frac {7}{4}+\frac {3 c_{4}}{2}\right )\right )\right )\right )^{\frac {3}{2}}} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = \frac {\sqrt {-12 c_{4}^{2}+28 c_{4} -11}\, \sqrt {3}}{6}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{4}\}\). Solving for the constants gives \begin {align*} c_{4}&=\frac {7}{6}-\frac {i \sqrt {5}}{3} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {x^{2}}{2}-\frac {4 \left (-1+\frac {3 \sin \left (x \right )}{2}+\frac {i \cos \left (x \right ) \sqrt {5}}{2}\right )^{2} \left (1+\frac {3 \sin \left (x \right )}{2}+\frac {i \cos \left (x \right ) \sqrt {5}}{2}\right )}{3 \left (\frac {3 \cos \left (x \right )}{2}-\frac {i \sin \left (x \right ) \sqrt {5}}{2}\right ) \sqrt {-\left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )+2\right ) \left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )-2\right )}}+\frac {7}{6}-\frac {i \sqrt {5}}{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{2} x^{2}+x +\frac {1}{2} \\ \tag{2} y &= \frac {7}{6}+\frac {x^{2}}{2}+\frac {\left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )-2\right )^{2} \left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )+2\right )}{\sqrt {2-3 i \sqrt {5}\, \sin \left (2 x \right )+7 \cos \left (2 x \right )}\, \left (3 i \sin \left (x \right ) \sqrt {5}-9 \cos \left (x \right )\right )}-\frac {i \sqrt {5}}{3} \\ \tag{3} y &= \frac {1}{2} x^{2}-x +\frac {1}{2} \\ \tag{4} y &= -\frac {2 \sin \left (x \right )}{3}+\frac {x^{2}}{2}+\frac {1}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{2} x^{2}+x +\frac {1}{2} \] Verified OK.
\[ y = \frac {7}{6}+\frac {x^{2}}{2}+\frac {\left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )-2\right )^{2} \left (i \cos \left (x \right ) \sqrt {5}+3 \sin \left (x \right )+2\right )}{\sqrt {2-3 i \sqrt {5}\, \sin \left (2 x \right )+7 \cos \left (2 x \right )}\, \left (3 i \sin \left (x \right ) \sqrt {5}-9 \cos \left (x \right )\right )}-\frac {i \sqrt {5}}{3} \] Warning, solution could not be verified
\[ y = \frac {1}{2} x^{2}-x +\frac {1}{2} \] Warning, solution could not be verified
\[ y = -\frac {2 \sin \left (x \right )}{3}+\frac {x^{2}}{2}+\frac {1}{2} \] Warning, solution could not be verified
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 3 *** 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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)+diff(y(x), x)-x, y(x)` *** Sublevel 4 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -_b(_a)+_a, _b(_a)` *** Sublevel 5 *** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful <- 2nd order ODE linearizable_by_differentiation successful ------------------- * Tackling next ODE. *** Sublevel 3 *** 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 <- 2nd order ODE linearizable_by_differentiation successful -> Calling odsolve with the ODE`, diff(y(x), x) = x+1, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful -> Calling odsolve with the ODE`, diff(y(x), x) = -1+x, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.391 (sec). Leaf size: 23
dsolve([diff(y(x),x$2)^2-2*diff(y(x),x$2)+diff(y(x),x)^2-2*x*diff(y(x),x)+x^2=0,y(0) = 1/2, D(y)(0) = 1],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\left (x +1\right )^{2}}{2} \\ y \left (x \right ) &= \frac {x^{2}}{2}+\sin \left (x \right )+\frac {1}{2} \\ \end{align*}
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{(y''[x])^2-2*y''[x]+(y'[x])^2-2*x*y'[x]+x^2==0,{y[0]==1/2,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
Not solved