1.24 problem 27

Internal problem ID [12887]
Internal file name [OUTPUT/11540_Monday_November_06_2023_01_33_09_PM_56463063/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 1.2. page 33
Problem number: 27.
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*} \left [y \left (0\right ) = {\frac {1}{2}}\right ] \end {align*}

1.24.1 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} \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

1.24.2 Solving as quadrature ode

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

Verification of solutions

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

1.24.3 Maple step by step solution

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

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: 9

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

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

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 10

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

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