1.27 problem 27 part(a)

Internal problem ID [4938]
Internal file name [OUTPUT/4431_Sunday_June_05_2022_01_20_44_PM_76753102/index.tex]

Book: Fundamentals of Differential Equations. By Nagle, Saff and Snider. 9th edition. Boston. Pearson 2018.
Section: Chapter 2, First order differential equations. Section 2.2, Separable Equations. Exercises. page 46
Problem number: 27 part(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]

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

1.27.1 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) &={\mathrm e}^{x^{2}} \end {align*}

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

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

1.27.2 Solving as quadrature ode

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

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

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2} \end {align*}

The constant \(c_{1} = 0\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2} \] Verified OK.

1.27.3 Maple step by step solution

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

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: 11

dsolve([diff(y(x),x)=exp(x^2),y(0) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\sqrt {\pi }\, \operatorname {erfi}\left (x \right )}{2} \]

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 16

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

\[ y(x)\to \frac {1}{2} \sqrt {\pi } \text {erfi}(x) \]