Internal problem ID [14137]
Internal file name [OUTPUT/13818_Saturday_March_02_2024_02_50_57_PM_91304232/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: 13 (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]
\[ \boxed {y^{\prime }-\sqrt {y^{2}-1}=0} \] With initial conditions \begin {align*} [y \left (4\right ) = -1] \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}-1} \end {align*}
The \(y\) domain of \(f(t,y)\) when \(t=4\) is \[ \{1\le y \le \infty , -\infty \le y \le -1\} \] And the point \(y_0 = -1\) 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}-1}\right ) \\ &= \frac {y}{\sqrt {y^{2}-1}} \end {align*}
The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=4\) is \[ \{-\infty \le y <-1, -1<y <1, 1<y \le \infty \} \] But the point \(y_0 = -1\) 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 \(y = -1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=-1 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -1 \\ \end{align*}
Verification of solutions
\[ y = -1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sqrt {y^{2}-1}=0, y \left (4\right )=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {y^{2}-1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y^{2}-1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y^{2}-1}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y+\sqrt {y^{2}-1}\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left ({\mathrm e}^{t +c_{1}}\right )^{2}+1}{2 \,{\mathrm e}^{t +c_{1}}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (4\right )=-1 \\ {} & {} & -1=\frac {\left ({\mathrm e}^{4+c_{1}}\right )^{2}+1}{2 \,{\mathrm e}^{4+c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\mathrm {I} \pi -4 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\mathrm {I} \pi -4\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {\left ({\mathrm e}^{2 t -8}+1\right ) {\mathrm e}^{4-t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {\left ({\mathrm e}^{2 t -8}+1\right ) {\mathrm e}^{4-t}}{2} \end {array} \]
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(y(t)^2-1),y(4) = -1],y(t), singsol=all)
\[ y \left (t \right ) = -1 \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 6
DSolve[{y'[t]==Sqrt[y[t]^2-1],{y[4]==-1}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to -1 \]