Internal problem ID [6118]
Internal file name [OUTPUT/5366_Sunday_June_05_2022_03_35_32_PM_75099697/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 1.2 THE NATURE OF SOLUTIONS.
Page 9
Problem number: 1(o).
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]
\[ \boxed {y^{2}+y^{\prime } y^{2}=-1} \]
Integrating both sides gives \begin {align*} \int -\frac {y^{2}}{y^{2}+1}d y &= \int {dx}\\ \int _{}^{y}-\frac {\textit {\_a}^{2}}{\textit {\_a}^{2}+1}d \textit {\_a}&= x +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}-\frac {\textit {\_a}^{2}}{\textit {\_a}^{2}+1}d \textit {\_a} &= x +c_{1} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}-\frac {\textit {\_a}^{2}}{\textit {\_a}^{2}+1}d \textit {\_a} = x +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2}+y^{\prime } y^{2}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-1-y^{2}}{y^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y^{2}}{-1-y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y^{2}}{-1-y^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -y+\arctan \left (y\right )=x +c_{1} \end {array} \]
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: 0.031 (sec). Leaf size: 14
dsolve(1+y(x)^2+y(x)^2*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (\operatorname {RootOf}\left (-\textit {\_Z} +x +c_{1} +\tan \left (\textit {\_Z} \right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.186 (sec). Leaf size: 35
DSolve[1+y[x]^2+y[x]^2*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}[\text {$\#$1}-\arctan (\text {$\#$1})\&][-x+c_1] \\ y(x)\to -i \\ y(x)\to i \\ \end{align*}