Internal problem ID [9975]
Internal file name [OUTPUT/8922_Monday_June_06_2022_05_52_25_AM_87933215/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1654 (book 6.63).
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }-a \left ({y^{\prime }}^{2}+1\right )^{\frac {3}{2}}=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right )-a \left (p \left (x \right )^{2}+1\right )^{\frac {3}{2}} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin{align*} \int \frac {1}{a \left (p^{2}+1\right )^{\frac {3}{2}}}d p &= \int d x \\ \frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}\, a}&=x +c_{1} \\ \end{align*} 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 }}{\sqrt {{y^{\prime }}^{2}+1}\, a} = x +c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{1} \sqrt {-\frac {1}{c_{1}^{2} a^{2}+2 a^{2} c_{1} x +x^{2} a^{2}-1}}\, a +x \sqrt {-\frac {1}{c_{1}^{2} a^{2}+2 a^{2} c_{1} x +x^{2} a^{2}-1}}\, a\,\mathop {\mathrm {d}x}}\\ &= \frac {\sqrt {-\frac {1}{c_{1}^{2} a^{2}+2 a^{2} c_{1} x +x^{2} a^{2}-1}}\, \left (a c_{1} +x a +1\right ) \left (a c_{1} +x a -1\right )}{a}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {-\frac {1}{c_{1}^{2} a^{2}+2 a^{2} c_{1} x +x^{2} a^{2}-1}}\, \left (a c_{1} +x a +1\right ) \left (a c_{1} +x a -1\right )}{a}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {-\frac {1}{c_{1}^{2} a^{2}+2 a^{2} c_{1} x +x^{2} a^{2}-1}}\, \left (a c_{1} +x a +1\right ) \left (a c_{1} +x a -1\right )}{a}+c_{2} \] 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 ) = a \left (p \left (y \right )^{2}+1\right )^{\frac {3}{2}} \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin{align*} \int \frac {p}{a \left (p^{2}+1\right )^{\frac {3}{2}}}d p &= \int d y \\ -\frac {1}{\sqrt {p \left (y \right )^{2}+1}\, a}&=y +c_{1} \\ \end{align*} 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 {1}{\sqrt {{y^{\prime }}^{2}+1}\, a} = y+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 }&=\frac {\sqrt {1-c_{1}^{2} a^{2}-2 y c_{1} a^{2}-y^{2} a^{2}}}{\left (y+c_{1} \right ) a} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {1-c_{1}^{2} a^{2}-2 y c_{1} a^{2}-y^{2} a^{2}}}{\left (y+c_{1} \right ) 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 {\left (y +c_{1} \right ) a}{\sqrt {-c_{1}^{2} a^{2}-2 y c_{1} a^{2}-y^{2} a^{2}+1}}d y &= \int d x \\ -\frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {\left (y +c_{1} \right ) a}{\sqrt {-c_{1}^{2} a^{2}-2 y c_{1} a^{2}-y^{2} a^{2}+1}}d y &= \int d x \\ \frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a} &= x +c_{2} \\ \tag{2} \frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a} = x +c_{2} \] Verified OK.
\[ \frac {\sqrt {1-\left (y+c_{1} \right )^{2} a^{2}}}{a} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=a \left ({y^{\prime }}^{2}+1\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=a \left (u \left (x \right )^{2}+1\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=a \left (u \left (x \right )^{2}+1\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (u \left (x \right )^{2}+1\right )^{\frac {3}{2}}}=a \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (u \left (x \right )^{2}+1\right )^{\frac {3}{2}}}d x =\int a d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (x \right )}{\sqrt {u \left (x \right )^{2}+1}}=x a +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x a \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}+c_{1} \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=x a \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}+c_{1} \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=x a \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}+c_{1} \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \left (x a \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}+c_{1} \sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}\right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {-\frac {1}{x^{2} a^{2}+2 c_{1} a x +c_{1}^{2}-1}}\, \left (x a +c_{1} +1\right ) \left (x a +c_{1} -1\right )}{a}+c_{2} \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) = a*(_b(_a)^2+1)^(3/2), _b(_a), HINT = [[1, 0], [y, -_b^2-1]]` *** Sublevel 2 * symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0], [y, -_b^2-1]
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 59
dsolve(diff(diff(y(x),x),x)=a*(diff(y(x),x)^2+1)^(3/2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -i x +c_{1} \\ y \left (x \right ) &= i x +c_{1} \\ y \left (x \right ) &= \frac {\left (-1+\left (c_{1} +x \right )^{2} a^{2}\right ) \sqrt {-\frac {1}{-1+\left (c_{1} +x \right )^{2} a^{2}}}+c_{2} a}{a} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.835 (sec). Leaf size: 75
DSolve[-(a*(1 + y'[x]^2)^(3/2)) + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\frac {i \sqrt {a^2 x^2+2 a c_1 x-1+c_1{}^2}}{a} \\ y(x)\to \frac {i \sqrt {a^2 x^2+2 a c_1 x-1+c_1{}^2}}{a}+c_2 \\ \end{align*}