3.2 problem 2

Internal problem ID [11360]
Internal file name [OUTPUT/10343_Wednesday_May_17_2023_07_49_32_PM_21072012/index.tex]

Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.2 Antiderivatives. Exercises page 19
Problem number: 2.
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]

\[ \boxed {x^{\prime }=\frac {1+t}{\sqrt {t}}} \] With initial conditions \begin {align*} [x \left (1\right ) = 4] \end {align*}

3.2.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=0\\ q(t) &=\frac {1+t}{\sqrt {t}} \end {align*}

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

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

3.2.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} x &= \int { \frac {1+t}{\sqrt {t}}\,\mathop {\mathrm {d}t}}\\ &= \frac {2 \sqrt {t}\, \left (3+t \right )}{3}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=1\) and \(x=4\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 4 = \frac {8}{3}+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {4}{3}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {4}{3}} \end {align*}

Substituting this in the general solution gives \begin {align*} x&=2 \sqrt {t}+\frac {2 t^{\frac {3}{2}}}{3}+\frac {4}{3} \end {align*}

The constant \(c_{1} = {\frac {4}{3}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= 2 \sqrt {t}+\frac {2 t^{\frac {3}{2}}}{3}+\frac {4}{3} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ x = 2 \sqrt {t}+\frac {2 t^{\frac {3}{2}}}{3}+\frac {4}{3} \] Verified OK.

3.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime }=\frac {1+t}{\sqrt {t}}, x \left (1\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int x^{\prime }d t =\int \frac {1+t}{\sqrt {t}}d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & x=\frac {2 \sqrt {t}\, \left (3+t \right )}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & x=\frac {2 t^{\frac {3}{2}}}{3}+2 \sqrt {t}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} x \left (1\right )=4 \\ {} & {} & 4=\frac {8}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {4}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {4}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & x=2 \sqrt {t}+\frac {2 t^{\frac {3}{2}}}{3}+\frac {4}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=2 \sqrt {t}+\frac {2 t^{\frac {3}{2}}}{3}+\frac {4}{3} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 14

dsolve([diff(x(t),t)=(1+t)/sqrt(t),x(1) = 4],x(t), singsol=all)
 

\[ x \left (t \right ) = \frac {2 t^{\frac {3}{2}}}{3}+2 \sqrt {t}+\frac {4}{3} \]

✓ Solution by Mathematica

Time used: 0.019 (sec). Leaf size: 23

DSolve[{x'[t]==(1+t)/Sqrt[t],{x[1]==4}},x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \frac {2}{3} \left (t^{3/2}+3 \sqrt {t}+2\right ) \]