2.6 problem 1.1-3 (f)

Internal problem ID [2459]
Internal file name [OUTPUT/1951_Sunday_June_05_2022_02_40_37_AM_90808783/index.tex]

Book: Ordinary Differential Equations, Robert H. Martin, 1983
Section: Problem 1.1-3, page 6
Problem number: 1.1-3 (f).
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]

\[ \boxed {y^{\prime }=8 \,{\mathrm e}^{4 t}+t} \] With initial conditions \begin {align*} [y \left (0\right ) = 12] \end {align*}

2.6.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=0\\ q(t) &=8 \,{\mathrm e}^{4 t}+t \end {align*}

Hence the ode is \begin {align*} y^{\prime } = 8 \,{\mathrm e}^{4 t}+t \end {align*}

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

2.6.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { 8 \,{\mathrm e}^{4 t}+t\,\mathop {\mathrm {d}t}}\\ &= \frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=12\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 12 = 2+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 10 \end {align*}

Trying the constant \begin {align*} c_{1} = 10 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \end {align*}

The constant \(c_{1} = 10\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \] Verified OK.

2.6.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=8 \,{\mathrm e}^{4 t}+t , y \left (0\right )=12\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int y^{\prime }d t =\int \left (8 \,{\mathrm e}^{4 t}+t \right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=12 \\ {} & {} & 12=2+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =10 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =10\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \end {array} \]

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: 17

dsolve([diff(y(t),t)=8*exp(4*t)+t,y(0) = 12],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {t^{2}}{2}+2 \,{\mathrm e}^{4 t}+10 \]

✓ Solution by Mathematica

Time used: 0.011 (sec). Leaf size: 21

DSolve[{y'[t]==8*Exp[4*t]+t,y[0]==12},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{2} \left (t^2+4 e^{4 t}+20\right ) \]