Internal problem ID [11369]
Internal file name [OUTPUT/10352_Wednesday_May_17_2023_07_49_47_PM_12695813/index.tex
]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.3.1 Separable equations. Exercises
page 26
Problem number: 1(c).
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]
\[ \boxed {y^{\prime }-y^{2}=1} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}+1}d y &= t +c_{1}\\ \arctan \left (y \right )&=t +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\tan \left (t +c_{1} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \tan \left (t +c_{1} \right ) \\ \end{align*}
Verification of solutions
\[ y = \tan \left (t +c_{1} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1+y^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{1+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{1+y^{2}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \arctan \left (y\right )=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\tan \left (t +c_{1} \right ) \end {array} \]
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(diff(y(t),t)=1+y(t)^2,y(t), singsol=all)
\[ y \left (t \right ) = \tan \left (t +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.216 (sec). Leaf size: 24
DSolve[y'[t]==1+y[t]^2,y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to \tan (t+c_1) \\ y(t)\to -i \\ y(t)\to i \\ \end{align*}