8.29 problem 10 (c)

Internal problem ID [12726]
Internal file name [OUTPUT/11379_Friday_November_03_2023_06_31_32_AM_295468/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.4.4, page 115
Problem number: 10 (c).
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]

\[ \boxed {y^{\prime }-\sqrt {\left (y+2\right ) \left (y-1\right )}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -3] \end {align*}

8.29.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \sqrt {\left (y +2\right ) \left (y -1\right )} \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[ \{1\le y \le \infty , -\infty \le y \le -2\} \] And the point \(y_0 = -3\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (\sqrt {\left (y +2\right ) \left (y -1\right )}\right ) \\ &= \frac {2 y +1}{2 \sqrt {\left (y +2\right ) \left (y -1\right )}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[ \{-\infty \le y <-2, -2

8.29.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {\left (y +2\right ) \left (y -1\right )}}d y &= \int {dx}\\ \ln \left (\frac {1}{2}+y +\sqrt {y^{2}+y -2}\right )&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=-3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\ln \left (2\right )+i \pi = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -\ln \left (2\right )+i \pi \end {align*}

Trying the constant \begin {align*} c_{1} = -\ln \left (2\right )+i \pi \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \ln \left (\frac {1}{2}+y +\sqrt {y^{2}+y -2}\right ) = x -\ln \left (2\right )+i \pi \end {align*}

The constant \(c_{1} = -\ln \left (2\right )+i \pi \) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\ln \left (2\right )+\ln \left (1+2 y+2 \sqrt {\left (y+2\right ) \left (y-1\right )}\right ) &= x -\ln \left (2\right )+i \pi \\ \end{align*}

Verification of solutions

\[ -\ln \left (2\right )+\ln \left (1+2 y+2 \sqrt {\left (y+2\right ) \left (y-1\right )}\right ) = x -\ln \left (2\right )+i \pi \] Verified OK.

8.29.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sqrt {\left (y+2\right ) \left (y-1\right )}=0, y \left (0\right )=-3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {\left (y+2\right ) \left (y-1\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {\left (y+2\right ) \left (y-1\right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {\left (y+2\right ) \left (y-1\right )}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (\frac {1}{2}+y+\sqrt {-2+y^{2}+y}\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {4 \left ({\mathrm e}^{x +c_{1}}\right )^{2}-4 \,{\mathrm e}^{x +c_{1}}+9}{8 \,{\mathrm e}^{x +c_{1}}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-3 \\ {} & {} & -3=\frac {4 \left ({\mathrm e}^{c_{1}}\right )^{2}-4 \,{\mathrm e}^{c_{1}}+9}{8 \,{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\ln \left (\frac {9}{2}\right )+\mathrm {I} \pi , -\ln \left (2\right )+\mathrm {I} \pi \right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\left (\ln \left (\frac {9}{2}\right )+\mathrm {I} \pi , -\ln \left (2\right )+\mathrm {I} \pi \right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {9 \,{\mathrm e}^{x}}{4}-\frac {1}{2}-\frac {{\mathrm e}^{-x}}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {9 \,{\mathrm e}^{x}}{4}-\frac {1}{2}-\frac {{\mathrm e}^{-x}}{4} \end {array} \]

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: 0.234 (sec). Leaf size: 16

dsolve([diff(y(x),x)=sqrt( (y(x)+2)*( y(x)-1)),y(0) = -3],y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {{\mathrm e}^{x}}{4}-\frac {1}{2}-\frac {9 \,{\mathrm e}^{-x}}{4} \]

✓ Solution by Mathematica

Time used: 0.021 (sec). Leaf size: 23

DSolve[{y'[x]==Sqrt[ (y[x]+2)*( y[x]-1)],{y[0]==-3}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{4} \left (-9 e^{-x}-e^x-2\right ) \]