Internal problem ID [12943]
Internal file name [OUTPUT/11596_Tuesday_November_07_2023_11_32_27_PM_640812/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.5 page 71
Problem number: 7.
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]
\[ \boxed {y^{\prime }-y \left (y-1\right ) \left (y-3\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= y \left (y -1\right ) \left (y -3\right ) \end {align*}
The \(y\) domain of \(f(t,y)\) when \(t=0\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} \int \frac {1}{y \left (y -1\right ) \left (y -3\right )}d y &= \int {dt}\\ \frac {\ln \left (y -3\right )}{6}+\frac {\ln \left (y \right )}{3}-\frac {\ln \left (y -1\right )}{2}&= t +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=2\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} \frac {i \pi }{6}+\frac {\ln \left (2\right )}{3} = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = \frac {i \pi }{6}+\frac {\ln \left (2\right )}{3} \end {align*}
Trying the constant \begin {align*} c_{1} = \frac {i \pi }{6}+\frac {\ln \left (2\right )}{3} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -3\right )}{6}+\frac {\ln \left (y \right )}{3}-\frac {\ln \left (y -1\right )}{2} = t +\frac {i \pi }{6}+\frac {\ln \left (2\right )}{3} \end {align*}
The constant \(c_{1} = \frac {i \pi }{6}+\frac {\ln \left (2\right )}{3}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \frac {\ln \left (y-3\right )}{6}+\frac {\ln \left (y\right )}{3}-\frac {\ln \left (y-1\right )}{2} &= t +\frac {i \pi }{6}+\frac {\ln \left (2\right )}{3} \\
\end{align*} Verification of solutions
\[
\frac {\ln \left (y-3\right )}{6}+\frac {\ln \left (y\right )}{3}-\frac {\ln \left (y-1\right )}{2} = t +\frac {i \pi }{6}+\frac {\ln \left (2\right )}{3}
\] Verified OK. Maple trace
✓ Solution by Maple
Time used: 4.344 (sec). Leaf size: 147
\[
y \left (t \right ) = \frac {\left (16 \,{\mathrm e}^{6 t}+9\right ) \left (1+8 \,{\mathrm e}^{6 t}+4 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}\right )^{\frac {2}{3}}+\left (24 \,{\mathrm e}^{6 t}+12 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}+9\right ) \left (1+8 \,{\mathrm e}^{6 t}+4 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}\right )^{\frac {1}{3}}+48 \,{\mathrm e}^{6 t}+24 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}+9}{\left (16 \,{\mathrm e}^{6 t}+3\right ) \left (1+8 \,{\mathrm e}^{6 t}+4 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}\right )^{\frac {2}{3}}+\left (8 \,{\mathrm e}^{6 t}+4 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}+3\right ) \left (1+8 \,{\mathrm e}^{6 t}+4 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}\right )^{\frac {1}{3}}+16 \,{\mathrm e}^{6 t}+8 \sqrt {{\mathrm e}^{6 t}+4 \,{\mathrm e}^{12 t}}+3}
\]
✓ Solution by Mathematica
Time used: 0.091 (sec). Leaf size: 105
\[
y(t)\to \frac {\sqrt [3]{2 \sqrt {e^{6 t} \left (4 e^{6 t}+1\right )^3}+8 e^{6 t}+16 e^{12 t}+1}}{4 e^{6 t}+1}+\frac {1}{\sqrt [3]{2 \sqrt {e^{6 t} \left (4 e^{6 t}+1\right )^3}+8 e^{6 t}+16 e^{12 t}+1}}+1
\]
4.3.2 Solving as quadrature ode
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([diff(y(t),t)=y(t)*(y(t)-1)*(y(t)-3),y(0) = 2],y(t), singsol=all)
DSolve[{y'[t]==y[t]*(y[t]-1)*(y[t]-3),{y[0]==2}},y[t],t,IncludeSingularSolutions -> True]