Internal problem ID [12725]
Internal file name [OUTPUT/11378_Friday_November_03_2023_06_31_31_AM_61785116/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 2. The Initial Value Problem. Exercises 2.4.4, page 115
Problem number: 10 (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]
\[ \boxed {y^{\prime }-\sqrt {\left (y+2\right ) \left (y-1\right )}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \sqrt {\left (y +2\right ) \left (y -1\right )} \end {align*}
The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[ \{1\le y \le \infty , -\infty \le y \le -2\} \] 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 {\left (y +2\right ) \left (y -1\right )}\right ) \\ &= \frac {2 y +1}{2 \sqrt {\left (y +2\right ) \left (y -1\right )}} \end {align*}
The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[
\{-\infty \le y <-2, -2
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. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\sqrt {\left (y+2\right ) \left (y-1\right )}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {\left (y+2\right ) \left (y-1\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {\left (y+2\right ) \left (y-1\right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {\left (y+2\right ) \left (y-1\right )}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (\frac {1}{2}+y+\sqrt {-2+y^{2}+y}\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {4 \left ({\mathrm e}^{x +c_{1}}\right )^{2}-4 \,{\mathrm e}^{x +c_{1}}+9}{8 \,{\mathrm e}^{x +c_{1}}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {4 \left ({\mathrm e}^{c_{1}}\right )^{2}-4 \,{\mathrm e}^{c_{1}}+9}{8 \,{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (\frac {3}{2}\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (\frac {3}{2}\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{x}}{4}-\frac {1}{2}+\frac {3 \,{\mathrm e}^{-x}}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{x}}{4}-\frac {1}{2}+\frac {3 \,{\mathrm e}^{-x}}{4} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 5
\[
y \left (x \right ) = 1
\]
✓ Solution by Mathematica
Time used: 0.022 (sec). Leaf size: 23
\[
y(x)\to \frac {1}{4} \left (3 e^{-x}+3 e^x-2\right )
\]
8.28.2 Solving as quadrature ode
8.28.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([diff(y(x),x)=sqrt( (y(x)+2)*( y(x)-1)),y(0) = 1],y(x), singsol=all)
DSolve[{y'[x]==Sqrt[ (y[x]+2)*( y[x]-1)],{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]