2.13 problem 2(c)

Internal problem ID [6151]
Internal file name [OUTPUT/5399_Sunday_June_05_2022_03_36_17_PM_95195842/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(c).
ODE order: 1.
ODE degree: 1.

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

Maple gives the following as the ode type

[_separable]

\[ \boxed {\frac {y^{\prime }}{x^{2}+1}-\frac {x}{y}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 3] \end {align*}

2.13.1 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 {x \left (x^{2}+1\right )}{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

2.13.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x \left (x^{2}+1\right )}{y} \end {align*}

Where \(f(x)=x \left (x^{2}+1\right )\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= x \left (x^{2}+1\right ) \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {x \left (x^{2}+1\right ) \,d x} \\ \frac {y^{2}}{2}&=\frac {\left (x^{2}+1\right )^{2}}{4}+c_{1} \\ \end{align*} Which results in \begin{align*} y &= \frac {\sqrt {2 x^{4}+4 x^{2}+8 c_{1} +2}}{2} \\ y &= -\frac {\sqrt {2 x^{4}+4 x^{2}+8 c_{1} +2}}{2} \\ \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 {2+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 {2+2 c_{1}} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {7}{2}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {7}{2}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \end {align*}

The constant \(c_{1} = {\frac {7}{2}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \] Verified OK.

2.13.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {y^{\prime }}{x^{2}+1}-\frac {x}{y}=0, y \left (1\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x \left (x^{2}+1\right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } y=x \left (x^{2}+1\right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } yd x =\int x \left (x^{2}+1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{2}}{2}=\frac {\left (x^{2}+1\right )^{2}}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {2 x^{4}+4 x^{2}+8 c_{1} +2}}{2}, y=\frac {\sqrt {2 x^{4}+4 x^{2}+8 c_{1} +2}}{2}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=-\frac {\sqrt {8+8 c_{1}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=\frac {\sqrt {8+8 c_{1}}}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {7}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {7}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \end {array} \]

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: 0.062 (sec). Leaf size: 20

dsolve([diff(y(x),x)/(1+x^2)=x/y(x),y(1) = 3],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\sqrt {2 x^{4}+4 x^{2}+30}}{2} \]

✓ Solution by Mathematica

Time used: 0.104 (sec). Leaf size: 25

DSolve[{y'[x]/(1+x^2)==x/y[x],{y[1]==3}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {\sqrt {x^4+2 x^2+15}}{\sqrt {2}} \]