Internal problem ID [14141]
Internal file name [OUTPUT/13822_Saturday_March_02_2024_02_50_58_PM_52727080/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 (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]
\[ \boxed {y^{\prime }-\sqrt {25-y^{2}}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 5] \end {align*}
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 \(t=0\) 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 \(t=0\) is \[ \{-5<y <5\} \] But the point \(y_0 = 5\) is not inside this domain. Hence existence and uniqueness theorem does not apply. Solution exists but no guarantee that unique solution exists.
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*}
Verification of solutions
\[ y = 5 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sqrt {25-y^{2}}=0, y \left (0\right )=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {25-y^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {25-y^{2}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {25-y^{2}}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arcsin \left (\frac {y}{5}\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=5 \sin \left (t +c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=5 \\ {} & {} & 5=5 \sin \left (c_{1} \right ) \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\pi }{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\pi }{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=5 \cos \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=5 \cos \left (t \right ) \end {array} \]
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(0) = 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[0]==5}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to 5 \]