4.6 problem 13

Internal problem ID [12946]
Internal file name [OUTPUT/11599_Tuesday_November_07_2023_11_51_50_PM_48642611/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 1.5 page 71
Problem number: 13.
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*} [y \left (0\right ) = 1] \end {align*}

4.6.1 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

4.6.2 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 \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -{\frac {1}{2}} = c_{1} \end {align*}

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

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

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

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

Summary

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

Verification of solutions

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

4.6.3 Maple step by step solution

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

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: 0.016 (sec). Leaf size: 11

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

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

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 14

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

\[ y(t)\to \frac {1}{\sqrt {1-2 t}} \]