Internal problem ID [4847]
Internal file name [OUTPUT/4340_Sunday_June_05_2022_01_03_31_PM_49431190/index.tex]
Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley.
2006
Section: Chapter 8, Ordinary differential equations. Section 7. Other second-Order equations.
page 435
Problem number: 6.
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear]]
\[ \boxed {-\frac {y^{\prime \prime }}{\left (1+y^{\prime }\right )^{\frac {3}{2}}}=-k} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int -\frac {y^{\prime \prime }}{\left (1+y^{\prime }\right )^{\frac {3}{2}}}d x &= \int -k d x\\ \frac {2}{\sqrt {1+y^{\prime }}} = -x k + c_{1} \end {align*}
Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { -\frac {x^{2} k^{2}-2 c_{1} k x +c_{1}^{2}-4}{\left (-x k +c_{1} \right )^{2}}\,\mathop {\mathrm {d}x}}\\ &= -x -\frac {4}{k \left (x k -c_{1} \right )}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x -\frac {4}{k \left (x k -c_{1} \right )}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -x -\frac {4}{k \left (x k -c_{1} \right )}+c_{2} \] Verified OK.
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*} k -\frac {p^{\prime }\left (x \right )}{\left (p \left (x \right )+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}{k \left (p +1\right )^{\frac {3}{2}}}d p &= \int d x \\ -\frac {2}{\sqrt {p \left (x \right )+1}\, k}&=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 {2}{\sqrt {1+y^{\prime }}\, k} = x +c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {c_{1}^{2} k^{2}+2 c_{1} k^{2} x +x^{2} k^{2}-4}{k^{2} \left (x +c_{1} \right )^{2}}\,\mathop {\mathrm {d}x}}\\ &= -x -\frac {4}{k^{2} \left (x +c_{1} \right )}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x -\frac {4}{k^{2} \left (x +c_{1} \right )}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -x -\frac {4}{k^{2} \left (x +c_{1} \right )}+c_{2} \] 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*} -\frac {p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )}{\left (1+p \left (y \right )\right )^{\frac {3}{2}}} = -k \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin{align*} \int \frac {p}{k \left (1+p \right )^{\frac {3}{2}}}d p &= \int d y \\ \frac {2 p \left (y \right )+4}{\sqrt {1+p \left (y \right )}\, k}&=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 {2 y^{\prime }+4}{k \sqrt {1+y^{\prime }}} = 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 }&=-2+\frac {\left (\frac {c_{1} k}{4}+\frac {y k}{4}+\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{4}\right ) c_{1} k}{2}+\frac {\left (\frac {c_{1} k}{4}+\frac {y k}{4}+\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{4}\right ) y k}{2} \tag {1} \\ y^{\prime }&=-2+\frac {\left (\frac {c_{1} k}{4}+\frac {y k}{4}-\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{4}\right ) c_{1} k}{2}+\frac {\left (\frac {c_{1} k}{4}+\frac {y k}{4}-\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{4}\right ) y k}{2} \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}{-2+\frac {c_{1}^{2} k^{2}}{8}+\frac {y c_{1} k^{2}}{4}+\frac {c_{1} k \sqrt {c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-16}}{8}+\frac {y^{2} k^{2}}{8}+\frac {y k \sqrt {c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-16}}{8}}d y &= \int d x \\ -\frac {y}{2}+\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{2 k}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int \frac {1}{-2+\frac {c_{1}^{2} k^{2}}{8}+\frac {y c_{1} k^{2}}{4}-\frac {c_{1} k \sqrt {c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-16}}{8}+\frac {y^{2} k^{2}}{8}-\frac {y k \sqrt {c_{1}^{2} k^{2}+2 y c_{1} k^{2}+y^{2} k^{2}-16}}{8}}d y &= \int d x \\ -\frac {y}{2}-\frac {\sqrt {y^{2} k^{2}+2 y c_{1} k^{2}+c_{1}^{2} k^{2}-16}}{2 k}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c_{1}^{2} k^{2}-4 c_{2}^{2} k^{2}-8 c_{2} k^{2} x -4 x^{2} k^{2}-16}{2 k^{2} \left (c_{1} -2 c_{2} -2 x \right )} \\ \tag{2} y &= -\frac {c_{1}^{2} k^{2}-4 c_{3}^{2} k^{2}-8 c_{3} k^{2} x -4 x^{2} k^{2}-16}{2 k^{2} \left (c_{1} -2 c_{3} -2 x \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {c_{1}^{2} k^{2}-4 c_{2}^{2} k^{2}-8 c_{2} k^{2} x -4 x^{2} k^{2}-16}{2 k^{2} \left (c_{1} -2 c_{2} -2 x \right )} \] Verified OK.
\[ y = -\frac {c_{1}^{2} k^{2}-4 c_{3}^{2} k^{2}-8 c_{3} k^{2} x -4 x^{2} k^{2}-16}{2 k^{2} \left (c_{1} -2 c_{3} -2 x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -\frac {y^{\prime \prime }}{\left (1+y^{\prime }\right )^{\frac {3}{2}}}=-k \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & -\frac {u^{\prime }\left (x \right )}{\left (1+u \left (x \right )\right )^{\frac {3}{2}}}=-k \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (1+u \left (x \right )\right )^{\frac {3}{2}}}=k \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (1+u \left (x \right )\right )^{\frac {3}{2}}}d x =\int k d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {2}{\sqrt {1+u \left (x \right )}}=x k +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {x^{2} k^{2}+2 c_{1} k x +c_{1}^{2}-4}{\left (x k +c_{1} \right )^{2}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {x^{2} k^{2}+2 c_{1} k x +c_{1}^{2}-4}{\left (x k +c_{1} \right )^{2}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {x^{2} k^{2}+2 c_{1} k x +c_{1}^{2}-4}{\left (x k +c_{1} \right )^{2}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {x^{2} k^{2}+2 c_{1} k x +c_{1}^{2}-4}{\left (x k +c_{1} \right )^{2}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-x -\frac {4}{k \left (x k +c_{1} \right )}+c_{2} \end {array} \]
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) = k*(_b(_a)+1)^(3/2), _b(_a), HINT = [[1, 0], [_a, -2-2*_b]]` *** Sublevel 2 ** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0], [_a, -2-2*_b]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 19
dsolve(k=diff(y(x),x$2)*(1+ (diff(y(x),x)))^(-3/2),y(x), singsol=all)
\[ y \left (x \right ) = -x -\frac {4}{k^{2} \left (x +c_{1} \right )}+c_{2} \]
✓ Solution by Mathematica
Time used: 0.515 (sec). Leaf size: 75
DSolve[k==y''[x]*(1+ (y'[x])^2)^(-3/2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\frac {i \sqrt {k^2 x^2+2 c_1 k x-1+c_1{}^2}}{k} \\ y(x)\to \frac {i \sqrt {k^2 x^2+2 c_1 k x-1+c_1{}^2}}{k}+c_2 \\ \end{align*}