1.26 problem 3(b)

Internal problem ID [6130]
Internal file name [OUTPUT/5378_Sunday_June_05_2022_03_35_45_PM_64565389/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.2 THE NATURE OF SOLUTIONS. Page 9
Problem number: 3(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]

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

1.26.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=0\\ q(x) &=2 \cos \left (x \right ) \sin \left (x \right ) \end {align*}

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

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

1.26.2 Solving as quadrature ode

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

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

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

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

Substituting this in the general solution gives \begin {align*} y&=1+\sin \left (x \right )^{2} \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 1+\sin \left (x \right )^{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 1+\sin \left (x \right )^{2} \] Verified OK.

1.26.3 Maple step by step solution

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

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: 12

dsolve([diff(y(x),x)=2*sin(x)*cos(x),y(0) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {\cos \left (2 x \right )}{2}+\frac {3}{2} \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 17

DSolve[{y'[x]==2*Sin[x]*Cos[x],{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} (3-\cos (2 x)) \]