Internal problem ID [11427]
Internal file name [OUTPUT/10410_Thursday_May_18_2023_04_18_39_AM_39873401/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 1.4.1. Integrating factors. Exercises
page 41
Problem number: 15(e).
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]
\[ \boxed {x^{\prime }-a x-b x^{3}=0} \]
Integrating both sides gives \begin {align*} \int \frac {1}{b \,x^{3}+a x}d x &= t +c_{1}\\ \frac {2 \ln \left (x \right )-\ln \left (b \,x^{2}+a \right )}{2 a}&=t +c_{1} \end {align*}
Solving for \(x\) gives these solutions \begin {align*} x_1&=\frac {\sqrt {-\left (b \,{\mathrm e}^{2 a c_{1} +2 t a}-1\right ) a \,{\mathrm e}^{2 a c_{1} +2 t a}}}{b \,{\mathrm e}^{2 a c_{1} +2 t a}-1}\\ &=\frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1}\\ x_2&=-\frac {\sqrt {-\left (b \,{\mathrm e}^{2 a c_{1} +2 t a}-1\right ) a \,{\mathrm e}^{2 a c_{1} +2 t a}}}{b \,{\mathrm e}^{2 a c_{1} +2 t a}-1}\\ &=-\frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= \frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1} \\ \tag{2} x &= -\frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1} \\ \end{align*}
Verification of solutions
\[ x = \frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1} \] Verified OK.
\[ x = -\frac {\sqrt {-\left (b \,c_{1}^{2} {\mathrm e}^{2 t a}-1\right ) a \,c_{1}^{2} {\mathrm e}^{2 t a}}}{b \,c_{1}^{2} {\mathrm e}^{2 t a}-1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime }-a x-b x^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=a x+b x^{3} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{a x+b x^{3}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{a x+b x^{3}}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (b x^{2}+a \right )}{2 a}+\frac {\ln \left (x\right )}{a}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & \left \{x=\frac {\sqrt {-\left (b \,{\mathrm e}^{2 c_{1} a +2 t a}-1\right ) a \,{\mathrm e}^{2 c_{1} a +2 t a}}}{b \,{\mathrm e}^{2 c_{1} a +2 t a}-1}, x=-\frac {\sqrt {-\left (b \,{\mathrm e}^{2 c_{1} a +2 t a}-1\right ) a \,{\mathrm e}^{2 c_{1} a +2 t a}}}{b \,{\mathrm e}^{2 c_{1} a +2 t a}-1}\right \} \end {array} \]
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: 0.0 (sec). Leaf size: 70
dsolve(diff(x(t),t)=a*x(t)+b*x(t)^3,x(t), singsol=all)
\begin{align*} x \left (t \right ) &= \frac {\sqrt {\left (c_{1} a \,{\mathrm e}^{-2 a t}-b \right ) a}}{c_{1} a \,{\mathrm e}^{-2 a t}-b} \\ x \left (t \right ) &= -\frac {\sqrt {\left (c_{1} a \,{\mathrm e}^{-2 a t}-b \right ) a}}{c_{1} a \,{\mathrm e}^{-2 a t}-b} \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.841 (sec). Leaf size: 118
DSolve[x'[t]==a*x[t]+b*x[t]^3,x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -\frac {i \sqrt {a} e^{a (t+c_1)}}{\sqrt {-1+b e^{2 a (t+c_1)}}} \\ x(t)\to \frac {i \sqrt {a} e^{a (t+c_1)}}{\sqrt {-1+b e^{2 a (t+c_1)}}} \\ x(t)\to 0 \\ x(t)\to -\frac {i \sqrt {a}}{\sqrt {b}} \\ x(t)\to \frac {i \sqrt {a}}{\sqrt {b}} \\ \end{align*}