Internal problem ID [11318]
Internal file name [OUTPUT/10304_Tuesday_December_27_2022_04_06_07_AM_10455936/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than first.
Article 58. Independent variable absent. Page 135
Problem number: Ex 3.
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]]
\[ \boxed {2 y^{\prime \prime }-{\mathrm e}^{y}=0} \]
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*}
The solution(s) found are the following \begin{align*} \tag{1} -\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} &= x +c_{2} \\ \tag{2} \frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = x +c_{2} \] Verified OK.
\[ \frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+c_{1}}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}} = x +c_{3} \] 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*} 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 \[ \frac {p \left (y \right )^{2}}{2}-\frac {{\mathrm e}^{y}}{2}-c_{1} = 0 \] 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}-c_{1} = 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 }&=\sqrt {{\mathrm e}^{y}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {{\mathrm e}^{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}^{y}+2 c_{1}}}d y &= \int d x \\ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+2 c_{1}}\, \sqrt {2}}{2 \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}+2 c_{1}}}d y &= \int d x \\ \frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{y}+2 c_{1}}\, \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} 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 ) \\ \tag{2} y &= \ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{3} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{2} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \] Verified OK.
\[ y = \ln \left (2 \tanh \left (\frac {\sqrt {c_{1}}\, \left (x +c_{3} \right ) \sqrt {2}}{2}\right )^{2} c_{1} -2 c_{1} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 \frac {d}{d x}y^{\prime }={\mathrm e}^{y} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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 ) \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: 0.469 (sec). Leaf size: 20
dsolve(2*diff(y(x),x$2)=exp(y(x)),y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (\frac {\sec \left (\frac {c_{2} +x}{2 c_{1}}\right )^{2}}{c_{1}^{2}}\right ) \]
✓ Solution by Mathematica
Time used: 60.049 (sec). Leaf size: 30
DSolve[2*y''[x]==Exp[y[x]],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \log \left (-c_1 \text {sech}^2\left (\frac {1}{2} \sqrt {c_1 (x+c_2){}^2}\right )\right ) \]