Internal problem ID [12952]
Internal file name [OUTPUT/11605_Tuesday_November_07_2023_11_51_56_PM_10796876/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 1.6 page 89
Problem number: 1 and 13 (iii).
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]
\[ \boxed {y^{\prime }-3 y \left (y-2\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= 3 y \left (y -2\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}{3 y \left (y -2\right )}d y &= \int {dt}\\ \frac {\ln \left (y -2\right )}{6}-\frac {\ln \left (y \right )}{6}&= t +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=3\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} -\frac {\ln \left (3\right )}{6} = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = -\frac {\ln \left (3\right )}{6} \end {align*}
Trying the constant \begin {align*} c_{1} = -\frac {\ln \left (3\right )}{6} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y -2\right )}{6}-\frac {\ln \left (y \right )}{6} = t -\frac {\ln \left (3\right )}{6} \end {align*}
The constant \(c_{1} = -\frac {\ln \left (3\right )}{6}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \frac {\ln \left (y-2\right )}{6}-\frac {\ln \left (y\right )}{6} &= t -\frac {\ln \left (3\right )}{6} \\
\end{align*} Verification of solutions
\[
\frac {\ln \left (y-2\right )}{6}-\frac {\ln \left (y\right )}{6} = t -\frac {\ln \left (3\right )}{6}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-3 y \left (y-2\right )=0, y \left (0\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=3 y \left (y-2\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (y-2\right )}=3 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y \left (y-2\right )}d t =\int 3d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y-2\right )}{2}-\frac {\ln \left (y\right )}{2}=3 t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {2}{{\mathrm e}^{6 t +2 c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=-\frac {2}{{\mathrm e}^{2 c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\frac {\ln \left (3\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\frac {\ln \left (3\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {6}{{\mathrm e}^{6 t}-3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {6}{{\mathrm e}^{6 t}-3} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
\[
y \left (t \right ) = -\frac {6}{{\mathrm e}^{6 t}-3}
\]
✓ Solution by Mathematica
Time used: 0.011 (sec). Leaf size: 16
\[
y(t)\to -\frac {6}{e^{6 t}-3}
\]
5.3.2 Solving as quadrature ode
5.3.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
<- Bernoulli successful`
dsolve([diff(y(t),t)=3*y(t)*(y(t)-2),y(0) = 3],y(t), singsol=all)
DSolve[{y'[t]==3*y[t]*(y[t]-2),{y[0]==3}},y[t],t,IncludeSingularSolutions -> True]