4.4 problem 4

4.4.1 Solving as quadrature ode
4.4.2 Maple step by step solution

Internal problem ID [14161]
Internal file name [OUTPUT/13842_Saturday_March_02_2024_02_51_17_PM_73020799/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.2, page 39
Problem number: 4.
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]

\[ \boxed {y^{\prime }-\frac {1+y^{2}}{y}=0} \]

4.4.1 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {y}{y^{2}+1}d y &= t +c_{1}\\ \frac {\ln \left (y^{2}+1\right )}{2}&=t +c_{1} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\sqrt {-1+{\mathrm e}^{2 t +2 c_{1}}}\\ &=\sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}}\\ y_2&=-\sqrt {-1+{\mathrm e}^{2 t +2 c_{1}}}\\ &=-\sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}} \\ \tag{2} y &= -\sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}} \\ \end{align*}

Figure 128: Slope field plot

Verification of solutions

\[ y = \sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}} \] Verified OK.

\[ y = -\sqrt {-1+{\mathrm e}^{2 t} c_{1}^{2}} \] Verified OK.

4.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {1+y^{2}}{y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1+y^{2}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{1+y^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime } y}{1+y^{2}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (1+y^{2}\right )}{2}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\sqrt {-1+{\mathrm e}^{2 t +2 c_{1}}}, y=-\sqrt {-1+{\mathrm e}^{2 t +2 c_{1}}}\right \} \end {array} \]

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: 0.016 (sec). Leaf size: 29

dsolve(diff(y(t),t)=(1+y(t)^2)/y(t),y(t), singsol=all)
 

\begin{align*} y \left (t \right ) &= \sqrt {c_{1} {\mathrm e}^{2 t}-1} \\ y \left (t \right ) &= -\sqrt {c_{1} {\mathrm e}^{2 t}-1} \\ \end{align*}

Solution by Mathematica

Time used: 2.441 (sec). Leaf size: 53

DSolve[y'[t]==(1+y[t]^2)/y[t],y[t],t,IncludeSingularSolutions -> True]
 

\begin{align*} y(t)\to -\sqrt {-1+e^{2 (t+c_1)}} \\ y(t)\to \sqrt {-1+e^{2 (t+c_1)}} \\ y(t)\to -i \\ y(t)\to i \\ \end{align*}