Internal problem ID [6246]
Internal file name [OUTPUT/5494_Sunday_June_05_2022_03_41_29_PM_83387578/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.9. Reduction of Order. Page
38
Problem number: 2(c).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-y^{\prime } {\mathrm e}^{y}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 2] \end {align*}
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-y^{\prime } {\mathrm e}^{y}\right )d x &= 0 \\ -{\mathrm e}^{y}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{{\mathrm e}^{y}+c_{1}}d y &= x +c_{2}\\ -\frac {-y +\ln \left ({\mathrm e}^{y}+c_{1} \right )}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} c_{1} +c_{1} x +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}\right ) \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*} 0 = c_{2} c_{1} +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\right )\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}+c_{1} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = -\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=-\ln \left (2\right ) \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \] 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 ) {\mathrm e}^{y} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { {\mathrm e}^{y}\,\mathop {\mathrm {d}y}}\\ &= {\mathrm e}^{y}+c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(y=0\) and \(p=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = 1+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 1 \end {align*}
Trying the constant \begin {align*} c_{1} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=1+{\mathrm e}^{y} \end {align*}
The constant \(c_{1} = 1\) 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*} y^{\prime } = 1+{\mathrm e}^{y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{1+{\mathrm e}^{y}}d y &= \int {dx}\\ y -\ln \left (1+{\mathrm e}^{y}\right )&= 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*} -\ln \left (2\right ) = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -\ln \left (2\right ) \end {align*}
Trying the constant \begin {align*} c_{2} = -\ln \left (2\right ) \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} y -\ln \left (1+{\mathrm e}^{y}\right ) = x -\ln \left (2\right ) \end {align*}
The constant \(c_{2} = -\ln \left (2\right )\) 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 (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \] Verified OK.
Writing the ode as \[ y^{\prime \prime }-y^{\prime } {\mathrm e}^{y} = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }-y^{\prime } {\mathrm e}^{y}\right )d x &= 0 \\ -{\mathrm e}^{y}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} \int \frac {1}{{\mathrm e}^{y}+c_{1}}d y &= x +c_{2}\\ -\frac {-y +\ln \left ({\mathrm e}^{y}+c_{1} \right )}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} c_{1} +c_{1} x +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}\right ) \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*} 0 = c_{2} c_{1} +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\right )\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}+c_{1} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = -\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=-\ln \left (2\right ) \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= 1\\ a_1 &= -{\mathrm e}^{y}\\ a_0 &= 0 \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1\,d y'} + \int {-{\mathrm e}^{y}\,d y} + \int {0\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} -{\mathrm e}^{y}+y^{\prime } = c_{1} \end {align*}
Which is now solved Integrating both sides gives \begin {align*} \int \frac {1}{{\mathrm e}^{y}+c_{1}}d y &= x +c_{2}\\ -\frac {-y +\ln \left ({\mathrm e}^{y}+c_{1} \right )}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{2} c_{1} +c_{1} x +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}\right ) \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*} 0 = c_{2} c_{1} +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\right )\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}+c_{1} \end {align*}
substituting \(y^{\prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = -\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1\\ c_{2}&=-\ln \left (2\right ) \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime } {\mathrm e}^{y}=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2\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 )-u \left (y \right ) {\mathrm e}^{y}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )={\mathrm e}^{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}u \left (y \right )\right )d y =\int {\mathrm e}^{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & u \left (y \right )={\mathrm e}^{y}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{y}+c_{1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{y}+c_{1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{y}+c_{1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{y}+c_{1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{y}+c_{1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{y}+c_{1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left ({\mathrm e}^{y}\right )}{c_{1}}-\frac {\ln \left ({\mathrm e}^{y}+c_{1} \right )}{c_{1}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=c_{2} c_{1} +c_{1} x +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{2} c_{1} +c_{1} x +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{2} c_{1} +\ln \left (-\frac {c_{1}}{-1+{\mathrm e}^{c_{2} c_{1}}}\right ) \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}}+c_{1} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1}}}{-1+{\mathrm e}^{c_{2} c_{1}}}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =1, c_{2} =-\ln \left (2\right )\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right )+x \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)-exp(_a)*_b(_a) = 0, _b(_a), HINT = [[1, _b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, _b]
✓ Solution by Maple
Time used: 0.735 (sec). Leaf size: 21
dsolve([diff(y(x),x$2)=diff(y(x),x)*exp(y(x)),y(0) = 0, D(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = x +\ln \left (-\frac {1}{{\mathrm e}^{x}-2}\right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y''[x]==y'[x]*Exp[y[x]],{y[0]==0,y'[0]==2}},y[x],x,IncludeSingularSolutions -> True]
{}