Internal problem ID [10067]
Internal file name [OUTPUT/9014_Monday_June_06_2022_06_13_07_AM_99188389/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1746 (book 6.155).
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]]
\[ \boxed {2 \left (y-a \right ) y^{\prime \prime }+{y^{\prime }}^{2}=-1} \]
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*} \left (2 y -2 a \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = -1 \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}+1}{2 \left (-y +a \right ) p} \end {align*}
Where \(f(y)=\frac {1}{-2 y +2 a}\) and \(g(p)=\frac {p^{2}+1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+1}{p}} \,dp &= \frac {1}{-2 y +2 a} \,d y \\ \int { \frac {1}{\frac {p^{2}+1}{p}} \,dp} &= \int {\frac {1}{-2 y +2 a} \,d y} \\ \frac {\ln \left (p^{2}+1\right )}{2}&=-\frac {\ln \left (-y +a \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+1} &= {\mathrm e}^{-\frac {\ln \left (-y +a \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}+1} &= \frac {c_{2}}{\sqrt {-y +a}} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{\sqrt {-y +a}} \] The solution is \[ \sqrt {p \left (y \right )^{2}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{\sqrt {-y +a}} \] 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}+1} = \frac {c_{2} {\mathrm e}^{c_{1}}}{\sqrt {-y+a}} \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 }&=-\frac {\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}{y-a} \tag {1} \\ y^{\prime }&=\frac {\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}{y-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 {y -a}{\sqrt {-\left (y -a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y -a \right )}}d y &= \int d x \\ \frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int \frac {y -a}{\sqrt {-\left (y -a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y -a \right )}}d y &= \int d x \\ -\frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}-\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2}&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2} &= x +c_{3} \\ \tag{2} -\frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}-\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2} = x +c_{3} \] Verified OK.
\[ -\frac {\left (-y+a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}{\sqrt {-\left (y-a \right ) \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}+y-a \right )}}-\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}} \arctan \left (\frac {y+\frac {c_{2}^{2} {\mathrm e}^{2 c_{1}}}{2}-a}{\sqrt {-y^{2}+\left (-c_{2}^{2} {\mathrm e}^{2 c_{1}}+2 a \right ) y+a \left (c_{2}^{2} {\mathrm e}^{2 c_{1}}-a \right )}}\right )}{2} = x +c_{4} \] Verified OK.
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)*(_b(_a)^2+1)/(_a-a) = 0, _b(_a), HINT = [[_a-a, 0]]` *** Subleve symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a-a, 0]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 123
dsolve(2*(y(x)-a)*diff(diff(y(x),x),x)+diff(y(x),x)^2+1=0,y(x), singsol=all)
\begin{align*} -\sqrt {-\left (-y \left (x \right )+a \right ) \left (-y \left (x \right )+c_{1} +a \right )}+\frac {c_{1} \arctan \left (\frac {2 y \left (x \right )-2 a -c_{1}}{2 \sqrt {-\left (-y \left (x \right )+a \right ) \left (-y \left (x \right )+c_{1} +a \right )}}\right )}{2}-x -c_{2} &= 0 \\ \sqrt {-\left (-y \left (x \right )+a \right ) \left (-y \left (x \right )+c_{1} +a \right )}-\frac {c_{1} \arctan \left (\frac {2 y \left (x \right )-2 a -c_{1}}{2 \sqrt {-\left (-y \left (x \right )+a \right ) \left (-y \left (x \right )+c_{1} +a \right )}}\right )}{2}-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.945 (sec). Leaf size: 595
DSolve[1 + y'[x]^2 + 2*(-a + y[x])*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {2} e^{2 c_1} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {2} e^{2 c_1} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {2} e^{2 (-c_1)} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 (-c_1)}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 (-c_1)}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {2} e^{2 (-c_1)} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 (-c_1)}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 (-c_1)}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {2} e^{2 c_1} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {2} e^{2 c_1} \arctan \left (\frac {\sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{\sqrt {2} \sqrt {a-\text {$\#$1}}}\right )+2 \sqrt {a-\text {$\#$1}} \sqrt {2 \text {$\#$1}-2 a+e^{2 c_1}}}{2 \sqrt {2}}\&\right ][x+c_2] \\ \end{align*}