3.2 problem 2
Internal
problem
ID
[7840]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
3.0
Problem
number
:
2
Date
solved
:
Monday, October 21, 2024 at 04:25:07 PM
CAS
classification
:
[_quadrature]
Solve
\begin{align*} w^{\prime }&=-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2} \end{align*}
With initial conditions
\begin{align*} w \left (1\right )&=-1 \end{align*}
3.2.1 Existence and uniqueness analysis
This is non linear first order ODE. In canonical form it is written as
\begin{align*} w^{\prime } &= f(z,w)\\ &= -\frac {1}{2}-\frac {\sqrt {1-12 w}}{2} \end{align*}
The \(w\) domain of \(f(z,w)\) when \(z=1\) is
\[
\left \{w \le \frac {1}{12}\right \}
\]
And the point \(w_0 = -1\) is inside this domain. Now we will look at the
continuity of
\begin{align*} \frac {\partial f}{\partial w} &= \frac {\partial }{\partial w}\left (-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}\right ) \\ &= \frac {3}{\sqrt {1-12 w}} \end{align*}
The \(w\) domain of \(\frac {\partial f}{\partial w}\) when \(z=1\) is
\[
\left \{w <\frac {1}{12}\right \}
\]
And the point \(w_0 = -1\) is inside this domain. Therefore solution exists and
is unique.
3.2.2 Solved as first order quadrature ode
Time used: 0.526 (sec)
Integrating gives
\begin{align*} \int \frac {1}{-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}}d w &= dz\\ \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (\sqrt {1-12 w}+1\right )}{3}&= z +c_1 \end{align*}
Solving for the constant of integration from initial conditions, the solution becomes
\begin{align*} \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (\sqrt {1-12 w}+1\right )}{3} = z -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (\sqrt {13}+1\right )}{3} \end{align*}
Figure 171: Slope field plot
\(w^{\prime } = -\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}\)
3.2.3 Solved as first order Exact ode
Time used: 0.192 (sec)
To solve an ode of the form
\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]
Hence
\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}
Comparing (A,B) shows
that
\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]
If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
\[ M(z,w) \mathop {\mathrm {d}z}+ N(z,w) \mathop {\mathrm {d}w}=0 \tag {1A} \]
Therefore
\begin{align*} \mathop {\mathrm {d}w} &= \left (-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}\right )\mathop {\mathrm {d}z}\\ \left (\frac {1}{2}+\frac {\sqrt {1-12 w}}{2}\right ) \mathop {\mathrm {d}z} + \mathop {\mathrm {d}w} &= 0 \tag {2A} \end{align*}
Comparing (1A) and (2A) shows that
\begin{align*} M(z,w) &= \frac {1}{2}+\frac {\sqrt {1-12 w}}{2}\\ N(z,w) &= 1 \end{align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the
following condition is satisfied
\[ \frac {\partial M}{\partial w} = \frac {\partial N}{\partial z} \]
Using result found above gives
\begin{align*} \frac {\partial M}{\partial w} &= \frac {\partial }{\partial w} \left (\frac {1}{2}+\frac {\sqrt {1-12 w}}{2}\right )\\ &= -\frac {3}{\sqrt {1-12 w}} \end{align*}
And
\begin{align*} \frac {\partial N}{\partial z} &= \frac {\partial }{\partial z} \left (1\right )\\ &= 0 \end{align*}
Since \(\frac {\partial M}{\partial w} \neq \frac {\partial N}{\partial z}\) , then the ODE is not exact . Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial w} - \frac {\partial N}{\partial z} \right ) \\ &=1\left ( \left ( -\frac {3}{\sqrt {1-12 w}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {3}{\sqrt {1-12 w}} \end{align*}
Since \(A\) depends on \(w\) , it can not be used to obtain an integrating factor. We will now try a
second method to find an integrating factor. Let
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial z} - \frac {\partial M}{\partial w} \right ) \\ &=\frac {2}{\sqrt {1-12 w}+1}\left ( \left ( 0\right ) - \left (-\frac {3}{\sqrt {1-12 w}} \right ) \right ) \\ &=\frac {6}{\left (\sqrt {1-12 w}+1\right ) \sqrt {1-12 w}} \end{align*}
Since \(B\) does not depend on \(z\) , it can be used to obtain an integrating factor. Let the
integrating factor be \(\mu \) . Then
\begin{align*} \mu &= e^{\int B \mathop {\mathrm {d}w}} \\ &= e^{\int \frac {6}{\left (\sqrt {1-12 w}+1\right ) \sqrt {1-12 w}}\mathop {\mathrm {d}w} } \end{align*}
The result of integrating gives
\begin{align*} \mu &= e^{-\ln \left (\sqrt {1-12 w}+1\right ) } \\ &= \frac {1}{\sqrt {1-12 w}+1} \end{align*}
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\)
and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\) .
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{\sqrt {1-12 w}+1}\left (\frac {1}{2}+\frac {\sqrt {1-12 w}}{2}\right ) \\ &= {\frac {1}{2}} \end{align*}
And
\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{\sqrt {1-12 w}+1}\left (1\right ) \\ &= \frac {1}{\sqrt {1-12 w}+1} \end{align*}
So now a modified ODE is obtained from the original ODE which will be exact and can be
solved using the standard method. The modified ODE is
\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}w}}{\mathop {\mathrm {d}z}} &= 0 \\ \left ({\frac {1}{2}}\right ) + \left (\frac {1}{\sqrt {1-12 w}+1}\right ) \frac { \mathop {\mathrm {d}w}}{\mathop {\mathrm {d}z}} &= 0 \end{align*}
The following equations are now set up to solve for the function \(\phi \left (z,w\right )\)
\begin{align*} \frac {\partial \phi }{\partial z } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial w } &= \overline {N}\tag {2} \end{align*}
Integrating (1) w.r.t. \(z\) gives
\begin{align*}
\int \frac {\partial \phi }{\partial z} \mathop {\mathrm {d}z} &= \int \overline {M}\mathop {\mathrm {d}z} \\
\int \frac {\partial \phi }{\partial z} \mathop {\mathrm {d}z} &= \int {\frac {1}{2}}\mathop {\mathrm {d}z} \\
\tag{3} \phi &= \frac {z}{2}+ f(w) \\
\end{align*}
Where \(f(w)\) is used for the constant of integration since \(\phi \) is a function
of both \(z\) and \(w\) . Taking derivative of equation (3) w.r.t \(w\) gives
\begin{equation}
\tag{4} \frac {\partial \phi }{\partial w} = 0+f'(w)
\end{equation}
But equation (2) says that \(\frac {\partial \phi }{\partial w} = \frac {1}{\sqrt {1-12 w}+1}\) .
Therefore equation (4) becomes
\begin{equation}
\tag{5} \frac {1}{\sqrt {1-12 w}+1} = 0+f'(w)
\end{equation}
Solving equation (5) for \( f'(w)\) gives
\[
f'(w) = \frac {1}{\sqrt {1-12 w}+1}
\]
Integrating the above w.r.t \(w\)
gives
\begin{align*}
\int f'(w) \mathop {\mathrm {d}w} &= \int \left ( \frac {1}{\sqrt {1-12 w}+1}\right ) \mathop {\mathrm {d}w} \\
f(w) &= -\frac {\sqrt {1-12 w}}{6}+\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}+ c_1 \\
\end{align*}
Where \(c_1\) is constant of integration. Substituting result found above for \(f(w)\) into
equation (3) gives \(\phi \)
\[
\phi = \frac {z}{2}-\frac {\sqrt {1-12 w}}{6}+\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}+ c_1
\]
But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new
constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = \frac {z}{2}-\frac {\sqrt {1-12 w}}{6}+\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}
\]
Solving for the constant of integration from initial conditions, the solution becomes
\begin{align*} \frac {z}{2}-\frac {\sqrt {1-12 w}}{6}+\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6} = \frac {1}{2}-\frac {\sqrt {13}}{6}+\frac {\ln \left (\sqrt {13}+1\right )}{6} \end{align*}
Figure 172: Slope field plot
\(w^{\prime } = -\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}\)
3.2.4 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [w^{\prime }=-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}, w \left (1\right )=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & w^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & w^{\prime }=-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {w^{\prime }}{-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} z \\ {} & {} & \int \frac {w^{\prime }}{-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}}d z =\int 1d z +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}-\frac {\ln \left (w\right )}{6}=z +\mathit {C1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} w \left (1\right )=-1 \\ {} & {} & \frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (\sqrt {13}+1\right )}{6}-\frac {\mathrm {I} \pi }{6}=1+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =-1+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (\sqrt {13}+1\right )}{6}-\frac {\mathrm {I} \pi }{6} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =-1+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (\sqrt {13}+1\right )}{6}-\frac {\mathrm {I} \pi }{6}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & \frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}-\frac {\ln \left (w\right )}{6}=z -1+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (\sqrt {13}+1\right )}{6}-\frac {\mathrm {I} \pi }{6} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & \frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (\sqrt {1-12 w}+1\right )}{6}-\frac {\ln \left (w\right )}{6}=z -1+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (\sqrt {13}+1\right )}{6}-\frac {\mathrm {I} \pi }{6} \end {array} \]
3.2.5 Maple trace
Methods for first order ODEs:
3.2.6 Maple dsolve solution
Solving time : 3.829
(sec)
Leaf size : 65
dsolve ([ diff ( w ( z ), z ) = -1/2-1/2*(1-12*w(z))^(1/2),
op ([ w (1) = -1])],w(z),singsol=all)
\[
w = \operatorname {RootOf}\left (6 z -2 \sqrt {1-12 \textit {\_Z}}-\ln \left (\sqrt {1-12 \textit {\_Z}}-1\right )+\ln \left (\sqrt {1-12 \textit {\_Z}}+1\right )+\ln \left (\textit {\_Z} \right )-6+2 \sqrt {13}+\ln \left (\sqrt {13}-1\right )-\ln \left (\sqrt {13}+1\right )-i \pi \right )
\]
3.2.7 Mathematica DSolve solution
Solving time : 11.607
(sec)
Leaf size : 105
DSolve [{ D [ w [ z ], z ] == -1/2 - Sqrt [1/4 - 3*w[z]],{w[1] == -1}},
w[z],z,IncludeSingularSolutions-> True ]
\begin{align*}
w(z)\to -\frac {1}{12} W\left (\left (\sqrt {13}-1\right ) e^{-3 z+\sqrt {13}+2}\right ) \left (W\left (\left (\sqrt {13}-1\right ) e^{-3 z+\sqrt {13}+2}\right )+2\right ) \\
w(z)\to -\frac {1}{12} W\left (\left (\sqrt {13}-1\right ) e^{-3 z+\sqrt {13}+2}\right ) \left (W\left (\left (\sqrt {13}-1\right ) e^{-3 z+\sqrt {13}+2}\right )+2\right ) \\
\end{align*}