8.8 problem 5 (a)

Internal problem ID [12705]
Internal file name [OUTPUT/11358_Friday_November_03_2023_06_31_03_AM_23906316/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.4.4, page 115
Problem number: 5 (a).
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]

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

8.8.1 Existence and uniqueness analysis

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

The \(y\) domain of \(f(x,y)\) when \(x=-1\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=-1\) is \[ \{-\infty

8.8.2 Solving as quadrature ode

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

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

Summary

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

Verification of solutions

\[ -\frac {1}{y} = x \] Verified OK.

8.8.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=0, y \left (-1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {1}{x +c_{1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=1 \\ {} & {} & 1=-\frac {1}{c_{1} -1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {1}{x} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {1}{x} \end {array} \]

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: 9

dsolve([diff(y(x),x)=y(x)^2,y(-1) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {1}{x} \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 10

DSolve[{y'[x]==y[x]^2,{y[-1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -\frac {1}{x} \]