4.49 problem 49

Internal problem ID [14206]
Internal file name [OUTPUT/13887_Saturday_March_09_2024_03_57_31_PM_31776992/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.2, page 39
Problem number: 49.
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]

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

4.49.1 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 {y}{\ln \left (y \right )} \end {align*}

The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[ \{0

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

4.49.2 Solving as quadrature ode

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

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

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

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

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

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

Summary

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

Verification of solutions

\[ \frac {\ln \left (y\right )^{2}}{2} = x +\frac {1}{2} \] Verified OK.

4.49.3 Maple step by step solution

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

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: 0.062 (sec). Leaf size: 12

dsolve([diff(y(x),x)=y(x)/ln(y(x)),y(0) = exp(1)],y(x), singsol=all)
 

\[ y \left (x \right ) = {\mathrm e}^{\sqrt {2 x +1}} \]

✓ Solution by Mathematica

Time used: 0.01 (sec). Leaf size: 16

DSolve[{y'[x]==y[x]/Log[y[x]],{y[0]==Exp[1]}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{\sqrt {2 x+1}} \]