2.11 problem 2(a)

Internal problem ID [6149]
Internal file name [OUTPUT/5397_Sunday_June_05_2022_03_36_14_PM_99221309/index.tex]

Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Section 1.3 SEPARABLE EQUATIONS. Page 12
Problem number: 2(a).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "separable", "differentialType", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_separable]

\[ \boxed {y y^{\prime }=1+x} \] With initial conditions \begin {align*} [y \left (1\right ) = 3] \end {align*}

2.11.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {1+x}{y} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=3\) is \[ \{-\infty

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=3\) is \[ \{-\infty

2.11.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {1+x}{y} \end {align*}

Where \(f(x)=1+x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= 1+x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {1+x \,d x} \\ \frac {y^{2}}{2}&=\frac {1}{2} x^{2}+x +c_{1} \\ \end{align*} Which results in \begin{align*} y &= \sqrt {x^{2}+2 c_{1} +2 x} \\ y &= -\sqrt {x^{2}+2 c_{1} +2 x} \\ \end{align*} Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 3 = -\sqrt {3+2 c_{1}} \end {align*}

Warning: Unable to solve for constant of integration. Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 3 = \sqrt {3+2 c_{1}} \end {align*}

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

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

Substituting this in the general solution gives \begin {align*} y&=\sqrt {x^{2}+2 x +6} \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {x^{2}+2 x +6} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \sqrt {x^{2}+2 x +6} \] Verified OK.

2.11.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y y^{\prime }=1+x , y \left (1\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y y^{\prime }d x =\int \left (1+x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\frac {1}{2} x^{2}+x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {x^{2}+2 c_{1} +2 x}, y=-\sqrt {x^{2}+2 c_{1} +2 x}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\sqrt {3+2 c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =3 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =3\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {x^{2}+2 x +6} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=-\sqrt {3+2 c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\sqrt {x^{2}+2 x +6} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.078 (sec). Leaf size: 14

dsolve([diff(y(x),x)*y(x)=x+1,y(1) = 3],y(x), singsol=all)
 

\[ y \left (x \right ) = \sqrt {x^{2}+2 x +6} \]

✓ Solution by Mathematica

Time used: 0.097 (sec). Leaf size: 17

DSolve[{y'[x]*y[x]==x+1,{y[1]==3}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \sqrt {x^2+2 x+6} \]