3.12 problem 12

Internal problem ID [14135]
Internal file name [OUTPUT/13816_Saturday_March_02_2024_02_50_56_PM_14765821/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.1, page 32
Problem number: 12.
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]

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

3.12.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=0\\ q(t) &=\frac {1}{t^{2}+1} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \frac {1}{t^{2}+1} \end {align*}

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

3.12.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) 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 this in the general solution gives \begin {align*} y&=\arctan \left (t \right ) \end {align*}

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

3.12.3 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 6

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

\[ y \left (t \right ) = \arctan \left (t \right ) \]

✓ Solution by Mathematica

Time used: 0.005 (sec). Leaf size: 7

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

\[ y(t)\to \arctan (t) \]