3.20 problem 14 (c)

Internal problem ID [14143]
Internal file name [OUTPUT/13824_Saturday_March_02_2024_02_51_00_PM_60166378/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: 14 (c).
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]

\[ \boxed {y^{\prime }-\sqrt {25-y^{2}}=0} \] With initial conditions \begin {align*} [y \left (4\right ) = -5] \end {align*}

3.20.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= \sqrt {-y^{2}+25} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=4\) is \[ \{-5\le y \le 5\} \] And the point \(y_0 = -5\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (\sqrt {-y^{2}+25}\right ) \\ &= -\frac {y}{\sqrt {-y^{2}+25}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=4\) is \[ \{-5

3.20.2 Solving as quadrature ode

Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = -5\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-5 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -5 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -5 \] Verified OK.

3.20.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sqrt {25-y^{2}}=0, y \left (4\right )=-5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {25-y^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {25-y^{2}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {25-y^{2}}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arcsin \left (\frac {y}{5}\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=5 \sin \left (t +c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (4\right )=-5 \\ {} & {} & -5=5 \sin \left (4+c_{1} \right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-4-\frac {\pi }{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-4-\frac {\pi }{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-5 \cos \left (t -4\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-5 \cos \left (t -4\right ) \end {array} \]

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: 0.0 (sec). Leaf size: 5

dsolve([diff(y(t),t)=sqrt(25-y(t)^2),y(4) = -5],y(t), singsol=all)
 

\[ y \left (t \right ) = -5 \]

✓ Solution by Mathematica

Time used: 0.001 (sec). Leaf size: 6

DSolve[{y'[t]==Sqrt[25-y[t]^2],{y[4]==-5}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -5 \]