2.16 problem Problem 16

Internal problem ID [2637]
Internal file name [OUTPUT/2129_Sunday_June_05_2022_02_49_40_AM_95114883/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Differential Equations. Section 1.4, Separable Differential Equations. page 43
Problem number: Problem 16.
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]

\[ \boxed {y^{\prime }-\frac {2 \sqrt {y-1}}{3}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1] \end {align*}

2.16.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {2 \sqrt {y -1}}{3} \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=1\) is \[ \{1\le y\} \] 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 (\frac {2 \sqrt {y -1}}{3}\right ) \\ &= \frac {1}{3 \sqrt {y -1}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=1\) is \[ \{1

2.16.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {2 \sqrt {y -1}}{3} \end {align*}

Where \(f(x)=1\) and \(g(y)=\frac {2 \sqrt {y -1}}{3}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {2 \sqrt {y -1}}{3}} \,dy &= 1 \,d x \\ \int { \frac {1}{\frac {2 \sqrt {y -1}}{3}} \,dy} &= \int {1 \,d x} \\ 3 \sqrt {y -1}&=x +c_{1} \\ \end{align*} The solution is \[ 3 \sqrt {y-1}-x -c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1-c_{1} = 0 \end {align*}

The solutions are \begin {align*} c_{1} = -1 \end {align*}

Trying the constant \begin {align*} c_{1} = -1 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} 3 \sqrt {y -1}-x +1 = 0 \end {align*}

The constant \(c_{1} = -1\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} 3 \sqrt {y-1}-x +1 &= 0 \\ \end{align*}

Verification of solutions

\[ 3 \sqrt {y-1}-x +1 = 0 \] Verified OK.

2.16.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {2 \sqrt {y-1}}{3}=0, y \left (1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 \sqrt {y-1}}{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y-1}}=\frac {2}{3} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y-1}}d x =\int \frac {2}{3}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 2 \sqrt {y-1}=\frac {2 x}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {1}{4} c_{1}^{2}+\frac {1}{3} c_{1} x +\frac {1}{9} x^{2}+1 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\frac {1}{4} c_{1}^{2}+\frac {1}{3} c_{1} +\frac {10}{9} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (-\frac {2}{3}, -\frac {2}{3}\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\left (-\frac {2}{3}, -\frac {2}{3}\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {1}{9} x^{2}-\frac {2}{9} x -\frac {2}{9} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {1}{9} x^{2}-\frac {2}{9} x -\frac {2}{9} \end {array} \]

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(x),x)=2/3*(y(x)-1)^(1/2),y(1) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = 1 \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 17

DSolve[{y'[x]==1/3*(y[x]-1)^(1/2),{y[1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{36} \left (x^2-2 x+37\right ) \]