1.18 problem 2(h)

Internal problem ID [3019]
Internal file name [OUTPUT/2511_Sunday_June_05_2022_03_17_23_AM_72583526/index.tex]

Book: Theory and solutions of Ordinary Differential equations, Donald Greenspan, 1960
Section: Exercises, page 14
Problem number: 2(h).
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]

\[ \boxed {y^{\prime }=x +\frac {1}{x}} \] With initial conditions \begin {align*} [y \left (-2\right ) = 5] \end {align*}

1.18.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=0\\ q(x) &=\frac {x^{2}+1}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \frac {x^{2}+1}{x} \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

1.18.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=-2\) and \(y=5\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 5 = 2+\ln \left (2\right )+i \pi +c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = -i \pi -\ln \left (2\right )+3 \end {align*}

Trying the constant \begin {align*} c_{1} = -i \pi -\ln \left (2\right )+3 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {x^{2}}{2}+\ln \left (x \right )-i \pi -\ln \left (2\right )+3 \end {align*}

The constant \(c_{1} = -i \pi -\ln \left (2\right )+3\) gives valid solution.

Summary

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

Verification of solutions

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

1.18.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=x +\frac {1}{x}, y \left (-2\right )=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (x +\frac {1}{x}\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2}+\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x^{2}}{2}+\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-2\right )=5 \\ {} & {} & 5=2+\ln \left (2\right )+\mathrm {I} \pi +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\mathrm {I} \pi -\ln \left (2\right )+3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\mathrm {I} \pi -\ln \left (2\right )+3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2}+\ln \left (x \right )-\mathrm {I} \pi -\ln \left (2\right )+3 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2}+\ln \left (x \right )-\mathrm {I} \pi -\ln \left (2\right )+3 \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

✓ Solution by Maple

Time used: 0.032 (sec). Leaf size: 21

dsolve([diff(y(x),x)=x+1/x,y(-2) = 5],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 25

DSolve[{y'[x]==x+1/x,y[-2]==5},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {x^2}{2}+\log \left (\frac {x}{2}\right )-i \pi +3 \]