2.11 problem 13

Internal problem ID [21]
Internal file name [OUTPUT/21_Sunday_June_05_2022_01_33_50_AM_13432084/index.tex]

Book: Differential equations and linear algebra, 3rd ed., Edwards and Penney
Section: Section 1.3. Slope fields and solution curves. Page 26
Problem number: 13.
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]

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

2.11.1 Existence and uniqueness analysis

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

The \(y\) domain of \(f(x,y)\) when \(x=0\) 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 (y^{\frac {1}{3}}\right ) \\ &= \frac {1}{3 y^{\frac {2}{3}}} \end {align*}

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

2.11.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{y^{\frac {1}{3}}}d y &= \int {dx}\\ \frac {3 y^{\frac {2}{3}}}{2}&= x +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} {\frac {3}{2}} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {3}{2}} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {3 y^{\frac {2}{3}}}{2} = x +\frac {3}{2} \end {align*}

The constant \(c_{1} = {\frac {3}{2}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {3 y^{\frac {2}{3}}}{2} &= x +\frac {3}{2} \\ \end{align*}

Verification of solutions

\[ \frac {3 y^{\frac {2}{3}}}{2} = x +\frac {3}{2} \] Verified OK.

2.11.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{\frac {1}{3}}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{\frac {1}{3}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{\frac {1}{3}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{\frac {1}{3}}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {3 y^{\frac {2}{3}}}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (6 x +6 c_{1} \right )^{\frac {3}{2}}}{27} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {2 \sqrt {6}\, c_{1}^{\frac {3}{2}}}{9} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {3}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {3}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 x +3\right ) \sqrt {6 x +9}}{9} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 x +3\right ) \sqrt {6 x +9}}{9} \end {array} \]

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: 13

dsolve([diff(y(x),x) = y(x)^(1/3),y(0) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (2 x +3\right ) \sqrt {6 x +9}}{9} \]

✓ Solution by Mathematica

Time used: 0.03 (sec). Leaf size: 23

DSolve[{y'[x] == y[x]^(1/3),y[0]==1},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {(2 x+3)^{3/2}}{3 \sqrt {3}} \]