1.7 problem 5.4 (ii)

Internal problem ID [11973]
Internal file name [OUTPUT/10626_Saturday_September_02_2023_02_48_37_PM_20629689/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 5, Trivial differential equations. Exercises page 33
Problem number: 5.4 (ii).
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]

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

1.7.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=0\\ q(x) &=x -\frac {1}{3} x^{3} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = x -\frac {1}{3} x^{3} \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

1.7.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { x -\frac {1}{3} x^{3}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (x^{2}-3\right )^{2}}{12}+c_{1} \end {align*}

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

The solutions are \begin {align*} c_{1} = {\frac {4}{3}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {4}{3}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\frac {1}{12} x^{4}+\frac {1}{2} x^{2}+\frac {7}{12} \end {align*}

The constant \(c_{1} = {\frac {4}{3}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{12} x^{4}+\frac {1}{2} x^{2}+\frac {7}{12} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {1}{12} x^{4}+\frac {1}{2} x^{2}+\frac {7}{12} \] Verified OK.

1.7.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=x -\frac {1}{3} x^{3}, y \left (-1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (x -\frac {1}{3} x^{3}\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-\frac {\left (x^{2}-3\right )^{2}}{12}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {1}{12} x^{4}+\frac {1}{2} x^{2}-\frac {3}{4}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=1 \\ {} & {} & 1=-\frac {1}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {4}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {4}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {1}{12} x^{4}+\frac {1}{2} x^{2}+\frac {7}{12} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {1}{12} x^{4}+\frac {1}{2} x^{2}+\frac {7}{12} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 15

dsolve([diff(y(x),x)=x-1/3*x^3,y(-1) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {\left (x^{2}-3\right )^{2}}{12}+\frac {4}{3} \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 21

DSolve[{y'[x]==x-1/3*x^3,{y[-1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{12} \left (-x^4+6 x^2+7\right ) \]