2.6.3 problem 3 (eq 41)
Internal
problem
ID
[18243]
Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929)
Section
:
Chapter
IV.
Methods
of
solution:
First
order
equations.
section
33.
Problems
at
page
91
Problem
number
:
3
(eq
41)
Date
solved
:
Monday, December 23, 2024 at 09:18:52 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} y^{\prime \prime }&=\frac {m \sqrt {1+{y^{\prime }}^{2}}}{k} \end{align*}
Solved as second order missing y ode
Time used: 0.320 (sec)
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 )-\frac {m \sqrt {1+p \left (x \right )^{2}}}{k} = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. Integrating gives
\begin{align*} \int \frac {k}{m \sqrt {p^{2}+1}}d p &= dx\\ \frac {k \,\operatorname {arcsinh}\left (p \right )}{m}&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} \frac {m \sqrt {p^{2}+1}}{k}&= 0 \end{align*}
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (x \right ) = -i\\ p \left (x \right ) = i \end{align*}
Solving for \(p \left (x \right )\) gives
\begin{align*}
p \left (x \right ) &= -i \\
p \left (x \right ) &= i \\
p \left (x \right ) &= \sinh \left (\frac {m \left (x +c_1 \right )}{k}\right ) \\
\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*} y^{\prime } = -i \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-i\, dx}\\ y &= -i x + c_2 \end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = i \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {i\, dx}\\ y &= i x + c_3 \end{align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = \sinh \left (\frac {m \left (x +c_1 \right )}{k}\right ) \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {\sinh \left (\frac {m \left (x +c_1 \right )}{k}\right )\, dx}\\ y &= \frac {k \cosh \left (\frac {m x}{k}+\frac {m c_1}{k}\right )}{m} + c_4 \end{align*}
\begin{align*} y&= \frac {k \cosh \left (\frac {m \left (x +c_1 \right )}{k}\right )}{m}+c_4 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= -i x +c_2 \\
y &= i x +c_3 \\
y &= \frac {k \cosh \left (\frac {m \left (x +c_1 \right )}{k}\right )}{m}+c_4 \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {m \sqrt {1+{y^{\prime }}^{2}}}{k} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {m \sqrt {1+u \left (x \right )^{2}}}{k} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {m \sqrt {1+u \left (x \right )^{2}}}{k} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\sqrt {1+u \left (x \right )^{2}}}=\frac {m}{k} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\sqrt {1+u \left (x \right )^{2}}}d x =\int \frac {m}{k}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arcsinh}\left (u \left (x \right )\right )=\frac {m x}{k}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (\frac {\mathit {C1} k +x m}{k}\right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sinh \left (\frac {\mathit {C1} k +x m}{k}\right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\sinh \left (\frac {\mathit {C1} k +x m}{k}\right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \sinh \left (\frac {\mathit {C1} k +x m}{k}\right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {k \cosh \left (\frac {m x}{k}+\mathit {C1} \right )}{m}+\mathit {C2} \end {array} \]
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)-m^2*(diff(y(x), x))/k^2, y(x)` *** Sublevel 2 ***
Methods for third order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
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) = m*(1+_b(_a)^2)^(1/2)/k, _b(_a), HINT = [[1, 0]]` *** Sublevel 2 ***
symmetry methods on request
`, `1st order, trying reduction of order with given symmetries:`[1, 0]
Maple dsolve solution
Solving time : 0.307
(sec)
Leaf size : 36
dsolve(diff(diff(y(x),x),x) = m/k*(1+diff(y(x),x)^2)^(1/2),
y(x),singsol=all)
\begin{align*}
y &= -i x +c_{1} \\
y &= i x +c_{1} \\
y &= c_{2} +\frac {k \cosh \left (\frac {m \left (x +c_{1} \right )}{k}\right )}{m} \\
\end{align*}
Mathematica DSolve solution
Solving time : 45.711
(sec)
Leaf size : 58
DSolve[{D[y[x],{x,2}]==m/k*Sqrt[1+D[y[x],x]^2],{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to c_2-\frac {k}{m \sqrt {\text {sech}^2\left (\frac {m x}{k}+c_1\right )}} \\
y(x)\to \frac {k}{m \sqrt {\text {sech}^2\left (\frac {m x}{k}+c_1\right )}}+c_2 \\
\end{align*}