Internal problem ID [2638]
Internal file name [OUTPUT/2130_Sunday_June_05_2022_02_49_44_AM_28871846/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 1, First-Order Differential Equations. Section 1.4, Separable Differential
Equations. page 43
Problem number: Problem 17.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {m v^{\prime }+k v^{2}=m g} \] With initial conditions \begin {align*} [v \left (0\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} v^{\prime } &= f(t,v)\\ &= -\frac {k \,v^{2}-m g}{m} \end {align*}
The \(v\) domain of \(f(t,v)\) when \(t=0\) is \[
\{-\infty The \(v\) domain of \(\frac {\partial f}{\partial v}\) when \(t=0\) is \[
\{-\infty
In canonical form the ODE is \begin {align*} v' &= F(t,v)\\ &= f( t) g(v)\\ &= \frac {-k \,v^{2}+m g}{m} \end {align*}
Where \(f(t)=1\) and \(g(v)=\frac {-k \,v^{2}+m g}{m}\). Integrating both sides gives \begin{align*}
\frac {1}{\frac {-k \,v^{2}+m g}{m}} \,dv &= 1 \,d t \\
\int { \frac {1}{\frac {-k \,v^{2}+m g}{m}} \,dv} &= \int {1 \,d t} \\
\frac {m \,\operatorname {arctanh}\left (\frac {k v}{\sqrt {m g k}}\right )}{\sqrt {m g k}}&=t +c_{1} \\
\end{align*} Which results in \begin{align*}
v &= \frac {\tanh \left (\frac {\sqrt {m g k}\, \left (t +c_{1} \right )}{m}\right ) \sqrt {m g k}}{k} \\
\end{align*} Initial conditions are used to
solve for \(c_{1}\). Substituting \(t=0\) and \(v=0\) in the above solution gives an equation to solve for the constant
of integration. \begin {align*} 0 = \frac {\sqrt {m g k}\, {\mathrm e}^{\frac {2 \sqrt {m g k}\, c_{1}}{m}}-\sqrt {m g k}}{k \,{\mathrm e}^{\frac {2 \sqrt {m g k}\, c_{1}}{m}}+k} \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} v&=\munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {\tanh \left (\frac {\sqrt {m g k}\, \left (t +c_{1} \right )}{m}\right ) \sqrt {m g k}}{k} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} v &= \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {\tanh \left (\frac {\sqrt {m g k}\, \left (t +c_{1} \right )}{m}\right ) \sqrt {m g k}}{k} \\
\end{align*} Verification of solutions
\[
v = \munderset {c_{1} \rightarrow 0}{\operatorname {lim}}\frac {\tanh \left (\frac {\sqrt {m g k}\, \left (t +c_{1} \right )}{m}\right ) \sqrt {m g k}}{k}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [m v^{\prime }+k v^{2}=m g , v \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & v^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & v^{\prime }=\frac {m g -k v^{2}}{m} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {v^{\prime }}{m g -k v^{2}}=\frac {1}{m} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {v^{\prime }}{m g -k v^{2}}d t =\int \frac {1}{m}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\mathrm {arctanh}\left (\frac {k v}{\sqrt {m g k}}\right )}{\sqrt {m g k}}=\frac {t}{m}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} v \\ {} & {} & v=\frac {\tanh \left (\frac {\sqrt {m g k}\, \left (c_{1} m +t \right )}{m}\right ) \sqrt {m g k}}{k} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} v \left (0\right )=0 \\ {} & {} & 0=\frac {\tanh \left (c_{1} \sqrt {m g k}\right ) \sqrt {m g k}}{k} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & v=\frac {\tanh \left (\frac {t \sqrt {m g k}}{m}\right ) \sqrt {m g k}}{k} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & v=\frac {\tanh \left (\frac {t \sqrt {m g k}}{m}\right ) \sqrt {m g k}}{k} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 26
\[
v \left (t \right ) = \frac {\tanh \left (\frac {\sqrt {m g k}\, t}{m}\right ) \sqrt {m g k}}{k}
\]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 39
\[
v(t)\to \frac {\sqrt {g} \sqrt {m} \tanh \left (\frac {\sqrt {g} \sqrt {k} t}{\sqrt {m}}\right )}{\sqrt {k}}
\]
2.17.2 Solving as separable ode
2.17.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([m*diff(v(t),t)=m*g-k*v(t)^2,v(0) = 0],v(t), singsol=all)
DSolve[{m*v'[t]==m*g-k*v[t]^2,{v[0]==0}},v[t],t,IncludeSingularSolutions -> True]