problem 40

Internal problem ID [2312]
Internal ﬁle name [OUTPUT/2312_Tuesday_February_27_2024_08_25_51_AM_92447401/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: 40.
ODE order: 2.
ODE degree: 1.

\[ \boxed {y y^{\prime \prime }-2 {y^{\prime }}^{2}-y^{2}=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = \sqrt {3}\right ] \end {align*}

18.40.1 Solving as second order ode missing x ode

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 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 p \left (y \right )^{2}-y^{2} = 0 \end {align*}

Which is now solved as ﬁrst order ode for \(p(y)\). Using the change of variables \(p \left (y \right ) = u \left (y \right ) y\) on the above ode results in new ode in \(u \left (y \right )\) \begin {align*} y^{2} u \left (y \right ) \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )-2 u \left (y \right )^{2} y^{2} = y^{2} \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= \frac {u^{2}+1}{u y} \end {align*}

Where \(f(y)=\frac {1}{y}\) and \(g(u)=\frac {u^{2}+1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+1}{u}} \,du &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {u^{2}+1}{u}} \,du} &= \int {\frac {1}{y} \,d y} \\ \frac {\ln \left (u^{2}+1\right )}{2}&=\ln \left (y \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}+1} &= {\mathrm e}^{\ln \left (y \right )+c_{2}} \end {align*}

Which simpliﬁes to \[ \sqrt {u \left (y \right )^{2}+1} = c_{3} {\mathrm e}^{c_{2}} y \] The solution is \[ \sqrt {u \left (y \right )^{2}+1} = c_{3} {\mathrm e}^{c_{2}} y \] Replacing \(u(y)\) in the above solution by \(\frac {p \left (y \right )}{y}\) results in the solution for \(p \left (y \right )\) in implicit form \begin {align*} \sqrt {\frac {p \left (y \right )^{2}}{y^{2}}+1} = c_{3} {\mathrm e}^{c_{2}} y\\ \sqrt {\frac {p \left (y \right )^{2}+y^{2}}{y^{2}}} = c_{3} {\mathrm e}^{c_{2}} y \end {align*}

Substituting initial conditions and solving for \(c_{2} \) gives \(c_{2} = \frac {\ln \left (\frac {4}{c_{3}^{2}}\right )}{2}\). Hence the solution becomes Initial conditions are used to solve for \(c_{3}\). Substituting \(y=1\) and \(p=\sqrt {3}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = 2 \sqrt {\frac {1}{c_{3}^{2}}}\, c_{3} \end {align*}

This solution is valid for any \(c_{3}\). Hence there are inﬁnite number of solutions.

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new ﬁrst order ode to solve which is \begin {align*} \sqrt {\frac {{y^{\prime }}^{2}+y^{2}}{y^{2}}} = 2 c_{3} \sqrt {\frac {1}{c_{3}^{2}}}\, y \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 }&=\sqrt {4 y^{2}-1}\, y \tag {1} \\ y^{\prime }&=-\sqrt {4 y^{2}-1}\, y \tag {2} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {4 y^{2}-1}\, y}d y &= \int {dx}\\ \arctan \left (\sqrt {4 y^{2}-1}\right )&= x +c_{4} \end {align*}

Initial conditions are used to solve for \(c_{4}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\pi }{3} = c_{4} \end {align*}

Substituting \(c_{4}\) found above in the general solution gives \begin {align*} \arctan \left (\sqrt {4 y^{2}-1}\right ) = x +\frac {\pi }{3} \end {align*}

Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {4 y^{2}-1}\, y}d y &= \int {dx}\\ \arctan \left (\frac {1}{\sqrt {4 y^{2}-1}}\right )&= x +c_{5} \end {align*}

Initial conditions are used to solve for \(c_{5}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\pi }{6} = c_{5} \end {align*}

Substituting \(c_{5}\) found above in the general solution gives \begin {align*} \arctan \left (\frac {1}{\sqrt {4 y^{2}-1}}\right ) = x +\frac {\pi }{6} \end {align*}

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.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \arctan \left (\sqrt {4 y^{2}-1}\right ) &= x +\frac {\pi }{3} \\ \tag{2} \arctan \left (\frac {1}{\sqrt {4 y^{2}-1}}\right ) &= x +\frac {\pi }{6} \\ \end{align*}

Veriﬁcation of solutions

\[ \arctan \left (\sqrt {4 y^{2}-1}\right ) = x +\frac {\pi }{3} \] Veriﬁed OK.

\[ \arctan \left (\frac {1}{\sqrt {4 y^{2}-1}}\right ) = x +\frac {\pi }{6} \] Warning, solution could not be veriﬁed

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 a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful`

\[ y \left (x \right ) = \frac {1}{-\sqrt {3}\, \sin \left (x \right )+\cos \left (x \right )} \]

18.40 problem 40

18.40.1 Solving as second order ode missing x ode