1.49 problem 56

Internal problem ID [14092]
Internal file name [OUTPUT/13773_Saturday_March_02_2024_02_49_25_PM_31005154/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 1. Introduction to Differential Equations. Exercises 1.1, page 10
Problem number: 56.
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]

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

1.49.1 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) &=\sin \left (2 t \right )-\cos \left (2 t \right ) \end {align*}

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

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

1.49.2 Solving as quadrature ode

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

1.49.3 Maple step by step solution

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

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: 18

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

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

✓ Solution by Mathematica

Time used: 0.044 (sec). Leaf size: 23

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

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