Internal problem ID [4669]
Internal file name [OUTPUT/4162_Sunday_June_05_2022_12_31_33_PM_72566539/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.19, page 504.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {2 y^{\prime \prime }-{\mathrm e}^{y}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ 2 y^{\prime } y^{\prime \prime }-y^{\prime } {\mathrm e}^{y} = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (2 y^{\prime } y^{\prime \prime }-y^{\prime } {\mathrm e}^{y}\right )d x &= 0 \\ {y^{\prime }}^{2}-{\mathrm e}^{y} = c_2 \end {align*}
Which is now solved for \(y\). 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 {{\mathrm e}^{y}+c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {{\mathrm e}^{y}+c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {{\mathrm e}^{y}+c_{1}}}d y &= \int d x \\ -\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {{\mathrm e}^{y}+c_{1}}}d y &= \int d x \\ \frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{3} \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} -\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = x +c_{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 0\) in the above gives \begin {align*} -\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {c_{1} +1}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\tanh \left (\frac {c_{2} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right ) c_{1}^{\frac {3}{2}} \left (1-\tanh \left (\frac {c_{2} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right )^{2}\right )}{\tanh \left (\frac {c_{2} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right )^{2} c_{1} -c_{1}} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = \frac {\left (-{\mathrm e}^{c_{2} \sqrt {c_{1}}}+1\right ) \sqrt {c_{1}}}{{\mathrm e}^{c_{2} \sqrt {c_{1}}}+1}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} \frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = x +c_{3} \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 \(y = 0\) and \(x = 0\) in the above gives \begin {align*} \frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {c_{1} +1}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \frac {\tanh \left (\frac {c_{3} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right ) c_{1}^{\frac {3}{2}} \left (1-\tanh \left (\frac {c_{3} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right )^{2}\right )}{\tanh \left (\frac {c_{3} \sqrt {c_{1}}}{2}+\frac {x \sqrt {c_{1}}}{2}\right )^{2} c_{1} -c_{1}} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = \frac {\left (-{\mathrm e}^{c_{3} \sqrt {c_{1}}}+1\right ) \sqrt {c_{1}}}{{\mathrm e}^{c_{3} \sqrt {c_{1}}}+1}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). There is no solution for the constants of integrations. This solution is removed.
Verification of solutions N/A
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*} 2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = {\mathrm e}^{y} \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 {{\mathrm e}^{y}}{2 p} \end {align*}
Where \(f(y)=\frac {{\mathrm e}^{y}}{2}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {{\mathrm e}^{y}}{2} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {{\mathrm e}^{y}}{2} \,d y} \\ \frac {p^{2}}{2}&=\frac {{\mathrm e}^{y}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {{\mathrm e}^{y}}{2}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=0\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \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*} \frac {p^{2}}{2}-\frac {{\mathrm e}^{y}}{2} = 0 \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*} \frac {{y^{\prime }}^{2}}{2}-\frac {{\mathrm e}^{y}}{2} = 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 }&={\mathrm e}^{\frac {y}{2}} \tag {1} \\ y^{\prime }&=-{\mathrm e}^{\frac {y}{2}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int {\mathrm e}^{-\frac {y}{2}}d y &= \int {dx}\\ -2 \,{\mathrm e}^{-\frac {y}{2}}&= x +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*} -2 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -2 \end {align*}
Trying the constant \begin {align*} c_{2} = -2 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -2 \,{\mathrm e}^{-\frac {y}{2}} = x -2 \end {align*}
The constant \(c_{2} = -2\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -{\mathrm e}^{-\frac {y}{2}}d y &= \int {dx}\\ 2 \,{\mathrm e}^{-\frac {y}{2}}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 2 \end {align*}
Trying the constant \begin {align*} c_{3} = 2 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} 2 \,{\mathrm e}^{-\frac {y}{2}} = x +2 \end {align*}
The constant \(c_{3} = 2\) 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 &= -2 \ln \left (1-\frac {x}{2}\right ) \\ \tag{2} y &= -2 \ln \left (\frac {x}{2}+1\right ) \\ \end{align*}
Verification of solutions
\[ y = -2 \ln \left (1-\frac {x}{2}\right ) \] Verified OK.
\[ y = -2 \ln \left (\frac {x}{2}+1\right ) \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y^{\prime \prime }={\mathrm e}^{y}, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 2 u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )={\mathrm e}^{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {{\mathrm e}^{y}}{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \frac {{\mathrm e}^{y}}{2}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=\frac {{\mathrm e}^{y}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {{\mathrm e}^{y}+2 c_{1}}, u \left (y \right )=-\sqrt {{\mathrm e}^{y}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{y}+2 c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{y}+2 c_{1}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {2}\, \mathrm {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {{\mathrm e}^{y}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{y}+2 c_{1}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{y}+2 c_{1}}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {2}\, \mathrm {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}}{2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right )}{2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-\frac {2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2}\right )}{2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2}\right )}{2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=\frac {2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2}\right )}{2 \tanh \left (\frac {\sqrt {c_{1}}\, c_{2} \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(1/2)*exp(_a) = 0, _b(_a), HINT = [[1, (1/2)*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 1/2*_b]
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 15
dsolve([2*diff(y(x),x$2)=exp(y(x)),y(0) = 0, D(y)(0) = 1],y(x), singsol=all)
\[ y \left (x \right ) = 2 \ln \left (2\right )+\ln \left (\frac {1}{\left (x -2\right )^{2}}\right ) \]
✓ Solution by Mathematica
Time used: 0.048 (sec). Leaf size: 15
DSolve[{2*y''[x]==Exp[y[x]],{y[0]==0,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -2 \log \left (1-\frac {x}{2}\right ) \]