1.23 problem 2.4 (a)

1.23.1 Existence and uniqueness analysis
1.23.2 Solving as quadrature ode
1.23.3 Maple step by step solution

Internal problem ID [13264]
Internal file name [OUTPUT/12436_Wednesday_February_14_2024_02_06_14_AM_1717262/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 2. Integration and differential equations. Additional exercises. page 32
Problem number: 2.4 (a).
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]

\[ \boxed {y^{\prime }=40 x \,{\mathrm e}^{2 x}} \] With initial conditions \begin {align*} [y \left (0\right ) = 4] \end {align*}

1.23.1 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) &=40 x \,{\mathrm e}^{2 x} \end {align*}

Hence the ode is \begin {align*} y^{\prime } = 40 x \,{\mathrm e}^{2 x} \end {align*}

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

1.23.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { 40 x \,{\mathrm e}^{2 x}\,\mathop {\mathrm {d}x}}\\ &= 10 \left (2 x -1\right ) {\mathrm e}^{2 x}+c_{1} \end {align*}

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=4\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 4 = -10+c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 14 \end {align*}

Trying the constant \begin {align*} c_{1} = 14 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=20 x \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}+14 \end {align*}

The constant \(c_{1} = 14\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 20 x \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}+14 \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 20 x \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}+14 \] Verified OK.

1.23.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=40 x \,{\mathrm e}^{2 x}, y \left (0\right )=4\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int 40 x \,{\mathrm e}^{2 x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=10 \left (2 x -1\right ) {\mathrm e}^{2 x}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=20 x \,{\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}+c_{1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=4 \\ {} & {} & 4=-10+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =14 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =14\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=14+\left (20 x -10\right ) {\mathrm e}^{2 x} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=14+\left (20 x -10\right ) {\mathrm e}^{2 x} \end {array} \]

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: 16

dsolve([diff(y(x),x)=4*x*10*exp(2*x),y(0) = 4],y(x), singsol=all)
 

\[ y \left (x \right ) = 14+\left (20 x -10\right ) {\mathrm e}^{2 x} \]

Solution by Mathematica

Time used: 0.014 (sec). Leaf size: 21

DSolve[{y'[x]==4*x*10*Exp[2*x],{y[0]==4}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to 2 \left (5 e^{2 x} (2 x-1)+7\right ) \]