3.8 problem 8

Internal problem ID [12935]
Internal file name [OUTPUT/11588_Tuesday_November_07_2023_11_27_32_PM_34093401/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: 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]

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

3.8.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= {\mathrm e}^{\frac {2}{y}} \end {align*}

The \(y\) domain of \(f(t,y)\) when \(t=1\) is \[ \{y <0\boldsymbol {\lor }0

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=1\) is \[ \{y <0\boldsymbol {\lor }0

3.8.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int {\mathrm e}^{-\frac {2}{y}}d y &= \int {dt}\\ y \,{\mathrm e}^{-\frac {2}{y}}-2 \,\operatorname {expIntegral}_{1}\left (\frac {2}{y}\right )&= t +c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=1\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) = 1+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) \end {align*}

Trying the constant \begin {align*} c_{1} = -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} y \,{\mathrm e}^{-\frac {2}{y}}-2 \,\operatorname {expIntegral}_{1}\left (\frac {2}{y}\right ) = t -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) \end {align*}

The constant \(c_{1} = -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right )\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y \,{\mathrm e}^{-\frac {2}{y}}-2 \,\operatorname {expIntegral}_{1}\left (\frac {2}{y}\right ) &= t -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) \\ \end{align*}

Verification of solutions

\[ y \,{\mathrm e}^{-\frac {2}{y}}-2 \,\operatorname {expIntegral}_{1}\left (\frac {2}{y}\right ) = t -1+2 \,{\mathrm e}^{-1}-2 \,\operatorname {expIntegral}_{1}\left (1\right ) \] Verified OK.

3.8.3 Maple step by step solution

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

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: 0.172 (sec). Leaf size: 38

dsolve([diff(y(t),t)=exp(2/y(t)),y(1) = 2],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {2}{\operatorname {RootOf}\left (2 \textit {\_Z} \,\operatorname {expIntegral}_{1}\left (1\right )-2 \textit {\_Z} \,\operatorname {expIntegral}_{1}\left (-\textit {\_Z} \right )-2 \textit {\_Z} \,{\mathrm e}^{-1}-t \textit {\_Z} -2 \,{\mathrm e}^{\textit {\_Z}}+\textit {\_Z} \right )} \]

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[{y'[t]==Exp[2/y[t]],{y[1]==2}},y[t],t,IncludeSingularSolutions -> True]
 

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