Internal problem ID [10050]
Internal file name [OUTPUT/8997_Monday_June_06_2022_06_10_17_AM_51630563/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1729 (book 6.138).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {2 y^{\prime \prime } y-{y^{\prime }}^{2}=-a} \]
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 ) y -p \left (y \right )^{2} = -a \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 {p^{2}-a}{2 p y} \end {align*}
Where \(f(y)=\frac {1}{2 y}\) and \(g(p)=\frac {p^{2}-a}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}-a}{p}} \,dp &= \frac {1}{2 y} \,d y \\ \int { \frac {1}{\frac {p^{2}-a}{p}} \,dp} &= \int {\frac {1}{2 y} \,d y} \\ \frac {\ln \left (p^{2}-a \right )}{2}&=\frac {\ln \left (y \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}-a} &= {\mathrm e}^{\frac {\ln \left (y \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}-a} &= c_{2} \sqrt {y} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}-a} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {p \left (y \right )^{2}-a} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}-a} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \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 {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a} \tag {1} \\ y^{\prime }&=-\sqrt {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a} \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 {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a}}d y &= \int d x \\ \frac {2 \sqrt {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a}\, {\mathrm e}^{-2 c_{1}}}{c_{2}^{2}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a}}d y &= \int d x \\ -\frac {2 \sqrt {y \,{\mathrm e}^{2 c_{1}} c_{2}^{2}+a}\, {\mathrm e}^{-2 c_{1}}}{c_{2}^{2}}&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{3} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +x^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}-4 a \right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \\ \tag{2} y &= \frac {\left (c_{4}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{4} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +x^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}-4 a \right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{3} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +x^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}-4 a \right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \] Verified OK.
\[ y = \frac {\left (c_{4}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{4} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +x^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}-4 a \right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 \left (\frac {d}{d x}y^{\prime }\right ) y-{y^{\prime }}^{2}=-a \\ \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 ) y -u \left (y \right )^{2}=-a \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )^{2}-a}{2 u \left (y \right ) y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}-a}=\frac {1}{2 y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}-a}d y =\int \frac {1}{2 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (y \right )^{2}-a \right )}{2}=\frac {\ln \left (y \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y \right )}}{{\mathrm e}^{-2 c_{1}}}, u \left (y \right )=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y \right )}}{{\mathrm e}^{-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y \right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}=\frac {1}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}d x =\int \frac {1}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 \sqrt {\left ({\mathrm e}^{-2 c_{1}}\right )^{2} a +{\mathrm e}^{-2 c_{1}} y}}{{\mathrm e}^{-2 c_{1}}}=\frac {x}{{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {c_{2}^{2} \left ({\mathrm e}^{-2 c_{1}}\right )^{2}-4 \left ({\mathrm e}^{-2 c_{1}}\right )^{2} a +2 c_{2} {\mathrm e}^{-2 c_{1}} x +x^{2}}{4 \,{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y \right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}=-\frac {1}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}} a +y\right )}}d x =\int -\frac {1}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 \sqrt {\left ({\mathrm e}^{-2 c_{1}}\right )^{2} a +{\mathrm e}^{-2 c_{1}} y}}{{\mathrm e}^{-2 c_{1}}}=-\frac {x}{{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {c_{2}^{2} \left ({\mathrm e}^{-2 c_{1}}\right )^{2}-4 \left ({\mathrm e}^{-2 c_{1}}\right )^{2} a -2 c_{2} {\mathrm e}^{-2 c_{1}} x +x^{2}}{4 \,{\mathrm e}^{-2 c_{1}}} \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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x), y(x)` *** Sublevel 2 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful <- 2nd order ODE linearizable_by_differentiation successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 24
dsolve(2*diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2+a=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (c_{1}^{2}-a \right ) x^{2}}{4 c_{2}}+c_{1} x +c_{2} \]
✓ Solution by Mathematica
Time used: 0.029 (sec). Leaf size: 36
DSolve[a - y'[x]^2 + 2*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^2 \left (-a+c_1{}^2\right )}{4 c_2}+c_1 x+c_2 \\ y(x)\to \text {Indeterminate} \\ \end{align*}