1.8 problem 8

Internal problem ID [5721]
Internal file name [OUTPUT/4969_Sunday_June_05_2022_03_15_29_PM_4200083/index.tex]

Book: Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold Scientific. Singapore. 1995
Section: Chapter 1. First order differential equations. Section 1.1 Separable equations problems. page 7
Problem number: 8.
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]

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

1.8.1 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)\\ &= 3 y^{\frac {2}{3}} \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=2\) is \[ \{0\le y\} \] And the point \(y_0 = 0\) is inside this domain. Now we will look at the continuity of \begin {align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (3 y^{\frac {2}{3}}\right ) \\ &= \frac {2}{y^{\frac {1}{3}}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=2\) is \[ \{0

1.8.2 Solving as separable ode

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

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

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} y^{\frac {1}{3}}-x +2 &= 0 \\ \end{align*}

Verification of solutions

\[ y^{\frac {1}{3}}-x +2 = 0 \] Verified OK.

1.8.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-3 y^{\frac {2}{3}}=0, y \left (2\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=3 y^{\frac {2}{3}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{\frac {2}{3}}}=3 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{\frac {2}{3}}}d x =\int 3d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & 3 y^{\frac {1}{3}}=3 x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x^{3}+c_{1} x^{2}+\frac {1}{3} c_{1}^{2} x +\frac {1}{27} c_{1}^{3} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (2\right )=0 \\ {} & {} & 0=8+4 c_{1} +\frac {2}{3} c_{1}^{2}+\frac {1}{27} c_{1}^{3} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (-6, -6, -6\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\left (-6, -6, -6\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (x -2\right )^{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (x -2\right )^{3} \end {array} \]

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: 0.0 (sec). Leaf size: 5

dsolve([diff(y(x),x)=3*y(x)^(2/3),y(2) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = 0 \]

✓ Solution by Mathematica

Time used: 0.002 (sec). Leaf size: 6

DSolve[{y'[x]==3*y[x]^(2/3),{y[2]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 0 \]