Internal problem ID [7192]
Internal file name [OUTPUT/6178_Sunday_June_05_2022_04_26_52_PM_3626605/index.tex]
Book: Own collection of miscellaneous problems
Section: section 3.0
Problem number: 2.
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]
\[ \boxed {w^{\prime }+\frac {\sqrt {1-12 w}}{2}=-{\frac {1}{2}}} \] With initial conditions \begin {align*} [w \left (1\right ) = -1] \end {align*}
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.
Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {1}{2}-\frac {\sqrt {1-12 w}}{2}}d w &= \int {dz}\\ \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{3}&= z +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(z=1\) and \(w=-1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} = 1+c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} \end {align*}
Trying the constant \begin {align*} c_{1} = -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{3} = z -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} \end {align*}
The constant \(c_{1} = -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{3} &= z -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {1-12 w}}{3}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{3} = z -1+\frac {\sqrt {13}}{3}-\frac {\ln \left (1+\sqrt {13}\right )}{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [w^{\prime }+\frac {\sqrt {1-12 w}}{2}=-\frac {1}{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 +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (w\right )}{6}+\frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{6}=z +c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} w \left (1\right )=-1 \\ {} & {} & -\frac {\mathrm {I} \pi }{6}+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (1+\sqrt {13}\right )}{6}=1+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-1-\frac {\mathrm {I} \pi }{6}+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (1+\sqrt {13}\right )}{6} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-1-\frac {\mathrm {I} \pi }{6}+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (1+\sqrt {13}\right )}{6}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (w\right )}{6}+\frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{6}=z -1-\frac {\mathrm {I} \pi }{6}+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (1+\sqrt {13}\right )}{6} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (w\right )}{6}+\frac {\sqrt {1-12 w}}{3}+\frac {\ln \left (\sqrt {1-12 w}-1\right )}{6}-\frac {\ln \left (1+\sqrt {1-12 w}\right )}{6}=z -1-\frac {\mathrm {I} \pi }{6}+\frac {\sqrt {13}}{3}+\frac {\ln \left (\sqrt {13}-1\right )}{6}-\frac {\ln \left (1+\sqrt {13}\right )}{6} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful`
✓ Solution by Maple
Time used: 0.64 (sec). Leaf size: 66
dsolve([diff(w(z),z) = -1/2 - sqrt(1/4 - 3*w(z)),w(1) = -1],w(z), singsol=all)
\[ w \left (z \right ) = \operatorname {RootOf}\left (-i \pi +2 \sqrt {13}-2 \sqrt {1-12 \textit {\_Z}}+\ln \left (\textit {\_Z} \right )-\ln \left (-1+\sqrt {1-12 \textit {\_Z}}\right )+\ln \left (1+\sqrt {1-12 \textit {\_Z}}\right )-\ln \left (1+\sqrt {13}\right )+\ln \left (-1+\sqrt {13}\right )+6 z -6\right ) \]
✓ Solution by Mathematica
Time used: 14.307 (sec). Leaf size: 105
DSolve[{w'[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*}