Internal problem ID [15182]
Internal file name [OUTPUT/15183_Tuesday_April_23_2024_04_53_39_PM_25005712/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 13. Basic concepts and definitions. Exercises
page 98
Problem number: 325.
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]]
\[ \boxed {y^{\prime \prime }-\left (1+{y^{\prime }}^{2}\right )^{\frac {3}{2}}=0} \]
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 )-\left (1+p \left (x \right )^{2}\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}{\left (p^{2}+1\right )^{\frac {3}{2}}}d p &= \int d x \\ \frac {p \left (x \right )}{\sqrt {1+p \left (x \right )^{2}}}&=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 {1+{y^{\prime }}^{2}}} = x +c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{1} \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}+x \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\,\mathop {\mathrm {d}x}}\\ &= \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\, \left (c_{1} +x +1\right ) \left (c_{1} +x -1\right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\, \left (c_{1} +x +1\right ) \left (c_{1} +x -1\right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\, \left (c_{1} +x +1\right ) \left (c_{1} +x -1\right )+c_{2} \] 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 ) = \left (1+p \left (y \right )^{2}\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}{\left (p^{2}+1\right )^{\frac {3}{2}}}d p &= \int d y \\ -\frac {1}{\sqrt {1+p \left (y \right )^{2}}}&=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 {1+{y^{\prime }}^{2}}} = 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 {-y^{2}-2 y c_{1} -c_{1}^{2}+1}}{y+c_{1}} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-y^{2}-2 y c_{1} -c_{1}^{2}+1}}{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 {y +c_{1}}{\sqrt {-c_{1}^{2}-2 c_{1} y -y^{2}+1}}d y &= \int d x \\ -\sqrt {-\left (y+c_{1} +1\right ) \left (y+c_{1} -1\right )}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {y +c_{1}}{\sqrt {-c_{1}^{2}-2 c_{1} y -y^{2}+1}}d y &= \int d x \\ \sqrt {-y^{2}-2 y c_{1} -c_{1}^{2}+1}&=c_{3} +x \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\sqrt {-\left (y+c_{1} +1\right ) \left (y+c_{1} -1\right )} &= x +c_{2} \\ \tag{2} \sqrt {-y^{2}-2 y c_{1} -c_{1}^{2}+1} &= c_{3} +x \\ \end{align*}
Verification of solutions
\[ -\sqrt {-\left (y+c_{1} +1\right ) \left (y+c_{1} -1\right )} = x +c_{2} \] Verified OK.
\[ \sqrt {-y^{2}-2 y c_{1} -c_{1}^{2}+1} = c_{3} +x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (1+{y^{\prime }}^{2}\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \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 )=\left (1+u \left (x \right )^{2}\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\left (1+u \left (x \right )^{2}\right )^{\frac {3}{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (1+u \left (x \right )^{2}\right )^{\frac {3}{2}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (1+u \left (x \right )^{2}\right )^{\frac {3}{2}}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (x \right )}{\sqrt {1+u \left (x \right )^{2}}}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=c_{1} \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}+x \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=c_{1} \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}+x \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=c_{1} \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}+x \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \left (c_{1} \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}+x \sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\sqrt {-\frac {1}{c_{1}^{2}+2 c_{1} x +x^{2}-1}}\, \left (c_{1} +x +1\right ) \left (c_{1} +x -1\right )+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) = (_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: 0.031 (sec). Leaf size: 49
dsolve(diff(y(x),x$2)=(1+diff(y(x),x)^2)^(3/2),y(x), singsol=all)
\begin{align*} y &= -i x +c_{1} \\ y &= i x +c_{1} \\ y &= \left (c_{1} +x +1\right ) \left (c_{1} +x -1\right ) \sqrt {-\frac {1}{\left (c_{1} +x +1\right ) \left (c_{1} +x -1\right )}}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.253 (sec). Leaf size: 59
DSolve[y''[x]==(1+y'[x]^2)^(3/2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-i \sqrt {x^2+2 c_1 x-1+c_1{}^2} \\ y(x)\to i \sqrt {x^2+2 c_1 x-1+c_1{}^2}+c_2 \\ \end{align*}