Internal problem ID [6838]
Internal file name [OUTPUT/6085_Thursday_July_28_2022_04_29_41_AM_88428700/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: 19.
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]]
\[ \boxed {y^{\prime \prime }+{\mathrm e}^{-2 y}=0} \] With initial conditions \begin {align*} [y \left (3\right ) = 0, y^{\prime }\left (3\right ) = -1] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }+y^{\prime } {\mathrm e}^{-2 y} = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }+y^{\prime } {\mathrm e}^{-2 y}\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {{\mathrm e}^{-2 y}}{2} = 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}^{-2 y}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {{\mathrm e}^{-2 y}+2 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}^{-2 y}+2 c_{1}}}d y &= \int d x \\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \sqrt {c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}}d y &= \int d x \\ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \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 {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \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 \(x = 3\) in the above gives \begin {align*} \frac {\operatorname {arctanh}\left (\frac {\sqrt {1+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right ) \sqrt {2}}{2 \sqrt {c_{1}}} = 3+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right )^{2}\right )}{2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1}} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 3\) in the above gives \begin {align*} -1 = \frac {\left ({\mathrm e}^{2 \left (3+c_{2} \right ) \sqrt {2}\, \sqrt {c_{1}}}-1\right ) \sqrt {2}\, \sqrt {c_{1}}}{{\mathrm e}^{2 \left (3+c_{2} \right ) \sqrt {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 {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \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 \(x = 3\) in the above gives \begin {align*} -\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right ) \sqrt {2}}{2 \sqrt {c_{1}}} = 3+c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{3} \right ) \sqrt {2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\sqrt {c_{1}}\, \left (x +c_{3} \right ) \sqrt {2}\right )^{2}\right )}{2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{3} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1}} \end {align*}
substituting \(y^{\prime } = -1\) and \(x = 3\) in the above gives \begin {align*} -1 = \frac {\left ({\mathrm e}^{2 \left (3+c_{3} \right ) \sqrt {2}\, \sqrt {c_{1}}}-1\right ) \sqrt {2}\, \sqrt {c_{1}}}{{\mathrm e}^{2 \left (3+c_{3} \right ) \sqrt {2}\, \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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = -{\mathrm e}^{-2 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}^{-2 y}}{p} \end {align*}
Where \(f(y)=-{\mathrm e}^{-2 y}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -{\mathrm e}^{-2 y} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-{\mathrm e}^{-2 y} \,d y} \\ \frac {p^{2}}{2}&=\frac {{\mathrm e}^{-2 y}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {{\mathrm e}^{-2 y}}{2}-c_{1} = 0 \] 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}^{-2 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}^{-2 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}^{-y} \tag {1} \\ y^{\prime }&=-{\mathrm e}^{-y} \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}^{y}d y &= \int {dx}\\ {\mathrm e}^{y}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=3\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = 3+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*} {\mathrm e}^{y} = x -2 \end {align*}
The constant \(c_{2} = -2\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int -{\mathrm e}^{y}d y &= \int {dx}\\ -{\mathrm e}^{y}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=3\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = 3+c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = -4 \end {align*}
Trying the constant \begin {align*} c_{3} = -4 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} -{\mathrm e}^{y} = x -4 \end {align*}
The constant \(c_{3} = -4\) 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 &= \ln \left (x -2\right ) \\ \tag{2} y &= \ln \left (-x +4\right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (x -2\right ) \] Warning, solution could not be verified
\[ y = \ln \left (-x +4\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }=-{\mathrm e}^{-2 y}, y \left (3\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}3\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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-{\mathrm e}^{-2 y} \\ \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 -{\mathrm e}^{-2 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=\frac {{\mathrm e}^{-2 y}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}, u \left (y \right )=-\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {{\mathrm e}^{-2 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}^{-2 y}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {{\mathrm e}^{-2 y}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 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}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \sqrt {c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {\ln \left (2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {{\mathrm e}^{-2 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}^{-2 y}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {{\mathrm e}^{-2 y}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 y}+2 c_{1}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 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}^{-2 y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{2 \sqrt {c_{1}}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {\ln \left (2 \tanh \left (\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\ln \left (2 \tanh \left (\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (3\right )=0 \\ {} & {} & 0=-\frac {\ln \left (2 \tanh \left (\sqrt {c_{1}}\, \left (-3+c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 \tanh \left (\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {2}\right )^{2}\right )}{2 \tanh \left (\sqrt {c_{1}}\, \left (-x +c_{2} \right ) \sqrt {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 {=}3\right \}}}}=-1 \\ {} & {} & -1=\frac {2 \tanh \left (\sqrt {c_{1}}\, \left (-3+c_{2} \right ) \sqrt {2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\sqrt {c_{1}}\, \left (-3+c_{2} \right ) \sqrt {2}\right )^{2}\right )}{2 \tanh \left (\sqrt {c_{1}}\, \left (-3+c_{2} \right ) \sqrt {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=-\frac {\ln \left (2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (3\right )=0 \\ {} & {} & 0=-\frac {\ln \left (2 \tanh \left (\left (3+c_{2} \right ) \sqrt {2}\, \sqrt {c_{1}}\right )^{2} c_{1} -2 c_{1} \right )}{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}\right )^{2}\right )}{2 \tanh \left (\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {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 {=}3\right \}}}}=-1 \\ {} & {} & -1=-\frac {2 \tanh \left (\left (3+c_{2} \right ) \sqrt {2}\, \sqrt {c_{1}}\right ) c_{1}^{\frac {3}{2}} \sqrt {2}\, \left (1-\tanh \left (\left (3+c_{2} \right ) \sqrt {2}\, \sqrt {c_{1}}\right )^{2}\right )}{2 \tanh \left (\left (3+c_{2} \right ) \sqrt {2}\, \sqrt {c_{1}}\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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 12
dsolve([diff(y(x),x$2)=-exp(-2*y(x)),y(3) = 0, D(y)(3) = -1],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\ln \left (\left (x -4\right )^{2}\right )}{2} \]
✓ Solution by Mathematica
Time used: 0.146 (sec). Leaf size: 11
DSolve[{y''[x]==-Exp[-2*y[x]],{y[3]==0,y'[3]==-1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \log (4-x) \]