5.28 problem 24

Internal problem ID [12977]
Internal file name [OUTPUT/11630_Tuesday_November_07_2023_11_52_51_PM_41929186/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.6 page 89
Problem number: 24.
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]

\[ \boxed {y^{\prime }-y^{2}+4 y=2} \] With initial conditions \begin {align*} [y \left (0\right ) = -2] \end {align*}

5.28.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= y^{2}-4 y +2 \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{-\infty

5.28.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}-4 y +2}d y &= \int {dt}\\ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 y -4\right ) \sqrt {2}}{4}\right )}{2}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=-2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 y -4\right ) \sqrt {2}}{4}\right )}{2} = t +\frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} \end {align*}

The constant \(c_{1} = \frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (y-2\right ) \sqrt {2}}{2}\right )}{2} &= t +\frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} \\ \end{align*}

Verification of solutions

\[ -\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (y-2\right ) \sqrt {2}}{2}\right )}{2} = t +\frac {\sqrt {2}\, \operatorname {arccoth}\left (2 \sqrt {2}\right )}{2}-\frac {i \pi \sqrt {2}}{4} \] Verified OK.

5.28.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}+4 y=2, y \left (0\right )=-2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-4 y+2 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}-4 y+2}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y^{2}-4 y+2}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {2}\, \mathrm {arctanh}\left (\frac {\left (2 y-4\right ) \sqrt {2}}{4}\right )}{2}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\left (\sqrt {2}-\tanh \left (\left (t +c_{1} \right ) \sqrt {2}\right )\right ) \sqrt {2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-2 \\ {} & {} & -2=\left (\sqrt {2}-\tanh \left (c_{1} \sqrt {2}\right )\right ) \sqrt {2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\sqrt {2}\, \mathrm {arctanh}\left (2 \sqrt {2}\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\sqrt {2}\, \mathrm {arctanh}\left (2 \sqrt {2}\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2-\sqrt {2}\, \tanh \left (\mathrm {arctanh}\left (2 \sqrt {2}\right )+\sqrt {2}\, t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2-\sqrt {2}\, \tanh \left (\mathrm {arctanh}\left (2 \sqrt {2}\right )+\sqrt {2}\, t \right ) \end {array} \]

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: 0.109 (sec). Leaf size: 24

dsolve([diff(y(t),t)=y(t)^2-4*y(t)+2,y(0) = -2],y(t), singsol=all)
 

\[ y \left (t \right ) = -\sqrt {2}\, \tanh \left (\operatorname {arctanh}\left (2 \sqrt {2}\right )+\sqrt {2}\, t \right )+2 \]

✓ Solution by Mathematica

Time used: 0.07 (sec). Leaf size: 59

DSolve[{y'[t]==y[t]^2-4*y[t]+2,{y[0]==-2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\frac {2 \left (\left (\sqrt {2}-3\right ) e^{2 \sqrt {2} t}+3+\sqrt {2}\right )}{\left (4+\sqrt {2}\right ) e^{2 \sqrt {2} t}-4+\sqrt {2}} \]