1.9 problem 5.4 (iv)

Internal problem ID [11975]
Internal file name [OUTPUT/10628_Saturday_September_02_2023_02_48_39_PM_65042289/index.tex]

Book: AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section: Chapter 5, Trivial differential equations. Exercises page 33
Problem number: 5.4 (iv).
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]

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

1.9.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} V^{\prime } + p(x)V &= 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*} V^{\prime } = \frac {x^{2}+1}{x} \end {align*}

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

1.9.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} V &= \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 \(x=1\) and \(V=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \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 this in the general solution gives \begin {align*} V&=\frac {x^{2}}{2}+\ln \left (x \right )+\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} V &= \frac {x^{2}}{2}+\ln \left (x \right )+\frac {1}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

1.9.3 Maple step by step solution

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

Maple trace

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

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 13

dsolve([x*diff(V(x),x)=1+x^2,V(1) = 1],V(x), singsol=all)
 

\[ V \left (x \right ) = \frac {x^{2}}{2}+\ln \left (x \right )+\frac {1}{2} \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 18

DSolve[{x*V'[x]==1+x^2,{V[1]==1}},V[x],x,IncludeSingularSolutions -> True]
 

\[ V(x)\to \frac {1}{2} \left (x^2+2 \log (x)+1\right ) \]