Internal problem ID [6097]
Internal file name [OUTPUT/5345_Sunday_June_05_2022_03_34_51_PM_97008130/index.tex
]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 6. Existence and uniqueness of solutions to systems and nth order equations.
Page 238
Problem number: 2.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-{y^{\prime }}^{2}=1} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \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*} p^{\prime }\left (x \right )-1-p \left (x \right )^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{p^{2}+1}d p &= \int {dx}\\ \arctan \left (p \right )&= x +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \arctan \left (p \right ) = x \end {align*}
The constant \(c_{1} = 0\) 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*} \arctan \left (y^{\prime }\right ) = x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \tan \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= -\ln \left (\cos \left (x \right )\right )+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\ln \left (\cos \left (x \right )\right ) \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{2} = 0\) 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.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\ln \left (\cos \left (x \right )\right ) \\ \end{align*}
Verification of solutions
\[ y = -\ln \left (\cos \left (x \right )\right ) \] Verified OK.
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{2} = 1 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int \frac {p}{p^{2}+1}d p &= \int {dy}\\ \frac {\ln \left (p^{2}+1\right )}{2}&= y +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(y=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 0 \end {align*}
Trying the constant \begin {align*} c_{3} = 0 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {\ln \left (p^{2}+1\right )}{2} = y \end {align*}
The constant \(c_{3} = 0\) 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 {\ln \left (1+{y^{\prime }}^{2}\right )}{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 {-1+{\mathrm e}^{2 y}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+{\mathrm e}^{2 y}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 0\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=0 \end {align*}
Solving equation (2)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 0\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=0 \end {align*}
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \end{align*}
Verification of solutions
\[ y = 0 \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-{y^{\prime }}^{2}=1, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )-u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right )^{2}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}+1}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arctan \left (u \left (x \right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \tan \left (x +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\ln \left (\tan \left (x +c_{1} \right )^{2}+1\right )}{2}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\frac {\ln \left (\tan \left (c_{1} \right )^{2}+1\right )}{2}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\tan \left (x +c_{1} \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=\tan \left (c_{1} \right ) \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (\sec \left (x \right )^{2}\right )}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\ln \left (\sec \left (x \right )^{2}\right )}{2} \end {array} \]
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`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 7
dsolve([diff(y(x),x$2)=1+diff(y(x),x)^2,y(0) = 0, D(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (\sec \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 2.581 (sec). Leaf size: 27
DSolve[{y''[x]==1+(y'[x])^2,{y[0]==0,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\log (-\cos (x))+i \pi \\ y(x)\to -\log (\cos (x)) \\ \end{align*}