8.32 problem 47

Internal problem ID [13059]
Internal file name [OUTPUT/11712_Wednesday_November_08_2023_03_29_10_AM_98042492/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. Review Exercises for chapter 1. page 136
Problem number: 47.
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]

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

8.32.1 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}+3 \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

8.32.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{-y^{2}+3}d y &= \int {dt}\\ \frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {y \sqrt {3}}{3}\right )}{3}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {y \sqrt {3}}{3}\right )}{3} = t \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {y \sqrt {3}}{3}\right )}{3} &= t \\ \end{align*}

Verification of solutions

\[ \frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {y \sqrt {3}}{3}\right )}{3} = t \] Verified OK.

8.32.3 Maple step by step solution

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

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: 0.032 (sec). Leaf size: 14

dsolve([diff(y(t),t)=3-y(t)^2,y(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = \sqrt {3}\, \tanh \left (\sqrt {3}\, t \right ) \]

✓ Solution by Mathematica

Time used: 0.047 (sec). Leaf size: 37

DSolve[{y'[t]==3-y[t]^2,{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {\sqrt {3} \left (e^{2 \sqrt {3} t}-1\right )}{e^{2 \sqrt {3} t}+1} \]