Internal problem ID [2308]
Internal file name [OUTPUT/2308_Tuesday_February_27_2024_08_25_48_AM_89056847/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 36.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {\left ({y^{\prime }}^{2}+1\right )^{2}-y^{2} y^{\prime \prime }=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = \sqrt {2}\right ] \end {align*}
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} -y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (p \left (y \right )^{3}+2 p \left (y \right )\right ) p \left (y \right ) = -1 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {\left (p^{2}+1\right )^{2}}{y^{2} p} \end {align*}
Where \(f(y)=\frac {1}{y^{2}}\) and \(g(p)=\frac {\left (p^{2}+1\right )^{2}}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (p^{2}+1\right )^{2}}{p}} \,dp &= \frac {1}{y^{2}} \,d y \\ \int { \frac {1}{\frac {\left (p^{2}+1\right )^{2}}{p}} \,dp} &= \int {\frac {1}{y^{2}} \,d y} \\ -\frac {1}{2 \left (p^{2}+1\right )}&=-\frac {1}{y}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 \left (p \left (y \right )^{2}+1\right )}+\frac {1}{y}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=3\) and \(p=\sqrt {2}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {1}{6}-c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = {\frac {1}{6}} \end {align*}
Trying the constant \begin {align*} c_{1} = {\frac {1}{6}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {p^{2} y -6 p^{2}+4 y -6}{6 \left (p^{2}+1\right ) y} = 0 \end {align*}
The constant \(c_{1} = {\frac {1}{6}}\) gives valid solution.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {\left (-y+6\right ) {y^{\prime }}^{2}-4 y+6}{6 y \left ({y^{\prime }}^{2}+1\right )} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {-2 \left (y-6\right ) \left (2 y-3\right )}}{y-6} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-2 \left (y-6\right ) \left (2 y-3\right )}}{y-6} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {y -6}{\sqrt {-2 \left (y -6\right ) \left (2 y -3\right )}}d y &= \int {dx}\\ -\frac {\sqrt {-4 y^{2}+30 y -36}}{4}-\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
Trying the constant \begin {align*} c_{2} = -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {\sqrt {-4 y^{2}+30 y -36}}{4}-\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} = x -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
The constant \(c_{2} = -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8}\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {y -6}{\sqrt {-2 \left (y -6\right ) \left (2 y -3\right )}}d y &= \int {dx}\\ \frac {\sqrt {-4 y^{2}+30 y -36}}{4}+\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = \frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
Trying the constant \begin {align*} c_{3} = \frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {\sqrt {-4 y^{2}+30 y -36}}{4}+\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} = x +\frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \end {align*}
The constant \(c_{3} = \frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8}\) 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} -\frac {\sqrt {-4 y^{2}+30 y-36}}{4}-\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} &= x -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \\ \tag{2} \frac {\sqrt {-4 y^{2}+30 y-36}}{4}+\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} &= x +\frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \\ \end{align*}
Verification of solutions
\[ -\frac {\sqrt {-4 y^{2}+30 y-36}}{4}-\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} = x -\frac {3 \sqrt {2}}{4}+\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \] Verified OK.
\[ \frac {\sqrt {-4 y^{2}+30 y-36}}{4}+\frac {9 \arcsin \left (\frac {4 y}{9}-\frac {5}{3}\right )}{8} = x +\frac {3 \sqrt {2}}{4}-\frac {9 \arcsin \left (\frac {1}{3}\right )}{8} \] 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 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(_b(_a)^2+1)^2/_a^2 = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 75.984 (sec). Leaf size: 221
dsolve([(1+diff(y(x),x)^2)^2=y(x)^2*diff(y(x),x$2),y(0) = 3, D(y)(0) = sqrt(2)],y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \operatorname {RootOf}\left (\sqrt {2}\, \left (\int _{\textit {\_Z}}^{3}\frac {\operatorname {RootOf}\left (\left (-\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} -1}{\sqrt {-\left (\operatorname {RootOf}\left (\left (-\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} -1\right ) \left (2 \operatorname {RootOf}\left (\left (-\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} +\textit {\_a} -2\right )}}d \textit {\_a} \right )+x \right ) \\ y \left (x \right ) &= \operatorname {RootOf}\left (\sqrt {2}\, \left (\int _{3}^{\textit {\_Z}}\frac {\operatorname {RootOf}\left (\left (\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} -1}{\sqrt {-\left (\operatorname {RootOf}\left (\left (\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} -1\right ) \left (2 \operatorname {RootOf}\left (\left (\sqrt {-\left (3 \textit {\_Z} -1\right ) \left (6 \textit {\_Z} +1\right )}+6 \textit {\_Z} -2\right ) \sqrt {2}\right ) \textit {\_a} +\textit {\_a} -2\right )}}d \textit {\_a} \right )+x \right ) \\ \end{align*}
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{(1+y'[x]^2)^2==y[x]^2*y''[x],{y[0]==3,y'[0]==Sqrt[2]}},y[x],x,IncludeSingularSolutions -> True]
Timed out