2.11 problem 11

Internal problem ID [4759]
Internal file name [OUTPUT/4252_Sunday_June_05_2022_12_48_14_PM_13953597/index.tex]

Book: Mathematical Methods in the Physical Sciences. third edition. Mary L. Boas. John Wiley. 2006
Section: Chapter 8, Ordinary differential equations. Section 2. Separable equations. page 398
Problem number: 11.
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]

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

The \(y\) domain of \(f(x,y)\) when \(x=1\) is \[ \{2\le y\} \] And the point \(y_0 = 3\) 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 {3 \left (y -2\right )^{\frac {1}{3}}}{2}\right ) \\ &= \frac {1}{2 \left (y -2\right )^{\frac {2}{3}}} \end {align*}

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=1\) is \[ \{2

2.11.2 Solving as separable ode

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

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

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \left (y -2\right )^{\frac {2}{3}}-x = 0 \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

Summary

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

Verification of solutions

\[ \left (y-2\right )^{\frac {2}{3}}-x = 0 \] Verified OK.

2.11.3 Maple step by step solution

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

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: 0.016 (sec). Leaf size: 9

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

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

✓ Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 12

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

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