3.34 problem 28

Internal problem ID [14157]
Internal file name [OUTPUT/13838_Saturday_March_02_2024_02_51_14_PM_98034743/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.1, page 32
Problem number: 28.
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]

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

3.34.1 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^{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

3.34.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int -\frac {1}{y^{3}}d y &= \int {dt}\\ \frac {1}{2 y^{2}}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y={\frac {1}{2}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 2 \end {align*}

Trying the constant \begin {align*} c_{1} = 2 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {1}{2 y^{2}} = 2+t \end {align*}

The constant \(c_{1} = 2\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {1}{2 y^{2}} &= 2+t \\ \end{align*}

Verification of solutions

\[ \frac {1}{2 y^{2}} = 2+t \] Verified OK.

3.34.3 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 11

dsolve([diff(y(t),t)=-y(t)^3,y(0) = 1/2],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {1}{\sqrt {2 t +4}} \]

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 18

DSolve[{y'[t]==-y[t]^3,{y[0]==1/2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {i}{\sqrt {-2 t-4}} \]