4.50 problem 50

Internal problem ID [14207]
Internal file name [OUTPUT/13888_Saturday_March_09_2024_03_57_31_PM_65770425/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.2, page 39
Problem number: 50.
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]

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

4.50.1 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) &=t \sin \left (t^{2}\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime } = t \sin \left (t^{2}\right ) \end {align*}

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

4.50.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=\sqrt {\pi }\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{1} +\frac {1}{2} \end {align*}

The solutions are \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-\frac {\cos \left (t^{2}\right )}{2}-\frac {1}{2} \end {align*}

The constant \(c_{1} = -{\frac {1}{2}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\cos \left (t^{2}\right )}{2}-\frac {1}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {\cos \left (t^{2}\right )}{2}-\frac {1}{2} \] Verified OK.

4.50.3 Maple step by step solution

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

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: 12

dsolve([diff(y(t),t)=t*sin(t^2),y(sqrt(Pi)) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {\cos \left (t^{2}\right )}{2}-\frac {1}{2} \]

✓ Solution by Mathematica

Time used: 0.013 (sec). Leaf size: 17

DSolve[{y'[t]==t*Sin[t^2],{y[Sqrt[Pi]]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\cos ^2\left (\frac {t^2}{2}\right ) \]