Internal problem ID [13450]
Internal file name [OUTPUT/12622_Friday_February_16_2024_12_00_25_AM_69897739/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 8. Review exercises for part of part II. page 143
Problem number: 29.
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]
\[ \boxed {y^{2}-y^{\prime }=-1} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+1}d y &= x +c_{1}\\ \arctan \left (y \right )&=x +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\tan \left (x +c_{1} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \tan \left (x +c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = \tan \left (x +c_{1} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2}-y^{\prime }=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1+y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{1+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{1+y^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arctan \left (y\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tan \left (x +c_{1} \right ) \end {array} \]
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: 0.0 (sec). Leaf size: 8
dsolve(y(x)^2+1-diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \tan \left (c_{1} +x \right ) \]
✓ Solution by Mathematica
Time used: 0.113 (sec). Leaf size: 24
DSolve[y[x]^2+1-y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \tan (x+c_1) \\ y(x)\to -i \\ y(x)\to i \\ \end{align*}