2.16 problem 2(e)

Internal problem ID [6154]
Internal file name [OUTPUT/5402_Sunday_June_05_2022_03_36_21_PM_69555050/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(e).
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]

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

2.16.1 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^{2}-1}{1+y} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=-2\) is \[ \{-\infty

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=-2\) is \[ \{-\infty

2.16.2 Solving as separable ode

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

Where \(f(x)=-x^{2}+1\) and \(g(y)=\frac {1}{1+y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{1+y}} \,dy &= -x^{2}+1 \,d x \\ \int { \frac {1}{\frac {1}{1+y}} \,dy} &= \int {-x^{2}+1 \,d x} \\ \frac {y \left (y +2\right )}{2}&=-\frac {1}{3} x^{3}+x +c_{1} \\ \end{align*} Which results in \begin{align*} y &= -1+\frac {\sqrt {-6 x^{3}+18 c_{1} +18 x +9}}{3} \\ y &= -1-\frac {\sqrt {-6 x^{3}+18 c_{1} +18 x +9}}{3} \\ \end{align*} Initial conditions are used to solve for \(c_{1}\). Substituting \(x=-1\) and \(y=-2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -2 = -1-\frac {\sqrt {-3+18 c_{1}}}{3} \end {align*}

The solutions are \begin {align*} c_{1} = {\frac {2}{3}} \end {align*}

Trying the constant \begin {align*} c_{1} = {\frac {2}{3}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=-1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \end {align*}

The constant \(c_{1} = {\frac {2}{3}}\) gives valid solution.

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=-1\) and \(y=-2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -2 = -1+\frac {\sqrt {-3+18 c_{1}}}{3} \end {align*}

Warning: Unable to solve for constant of integration.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \] Verified OK.

2.16.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime } \left (1+y\right )=-x^{2}+1, y \left (-1\right )=-2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \left (1+y\right )d x =\int \left (-x^{2}+1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y+\frac {y^{2}}{2}=-\frac {1}{3} x^{3}+x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-1-\frac {\sqrt {-6 x^{3}+18 c_{1} +18 x +9}}{3}, y=-1+\frac {\sqrt {-6 x^{3}+18 c_{1} +18 x +9}}{3}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=-2 \\ {} & {} & -2=-1-\frac {\sqrt {-3+18 c_{1}}}{3} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {2}{3} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {2}{3}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (-1\right )=-2 \\ {} & {} & -2=-1+\frac {\sqrt {-3+18 c_{1}}}{3} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \end {array} \]

Maple trace

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

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 20

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

\[ y \left (x \right ) = -1-\frac {\sqrt {-6 x^{3}+18 x +21}}{3} \]

✓ Solution by Mathematica

Time used: 0.151 (sec). Leaf size: 28

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

\[ y(x)\to -\frac {\sqrt {-2 x^3+6 x+7}}{\sqrt {3}}-1 \]