3.11 problem 15

Internal problem ID [12938]
Internal file name [OUTPUT/11591_Tuesday_November_07_2023_11_27_34_PM_56039897/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.4 page 61
Problem number: 15.
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]

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

3.11.1 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} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[ \{0\le y\} \] 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 (\sqrt {y}\right ) \\ &= \frac {1}{2 \sqrt {y}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[ \{0

3.11.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {y}}d y &= \int {dt}\\ 2 \sqrt {y}&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 2 \end {align*}

Trying the constant \begin {align*} c_{1} = 2 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} 2 \sqrt {y} = 2+t \end {align*}

The constant \(c_{1} = 2\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} 2 \sqrt {y} &= 2+t \\ \end{align*}

Verification of solutions

\[ 2 \sqrt {y} = 2+t \] Verified OK.

3.11.3 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 11

dsolve([diff(y(t),t)=sqrt( y(t)),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (t +2\right )^{2}}{4} \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 14

DSolve[{y'[t]==Sqrt[ y[t] ],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{4} (t+2)^2 \]