3.11 problem 14(b)

Internal problem ID [5943]
Internal file name [OUTPUT/5191_Sunday_June_05_2022_03_27_08_PM_70795525/index.tex]

Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY 1961
Section: Chapter 1. Introduction– Linear equations of First Order. Page 45
Problem number: 14(b).
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]

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

3.11.1 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}+1 \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[ \{-\infty

3.11.2 Solving as quadrature ode

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

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

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

Summary

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

Verification of solutions

\[ \arctan \left (y\right ) = x \] Verified OK.

3.11.3 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 6

dsolve([diff(y(x),x)=1+y(x)^2,y(0) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = \tan \left (x \right ) \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 7

DSolve[{y'[x]==1+y[x]^2,{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \tan (x) \]