Internal problem ID [533]
Internal file name [OUTPUT/533_Sunday_June_05_2022_01_43_11_AM_86686196/index.tex]
Book: Elementary differential equations and boundary value problems, 10th ed., Boyce and
DiPrima
Section: Section 2.5. Page 88
Problem number: 3.
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]
\[ \boxed {y^{\prime }-y \left (-2+y\right ) \left (-1+y\right )=0} \]
Integrating both sides gives \begin {align*} \int \frac {1}{y \left (y -2\right ) \left (y -1\right )}d y &= t +c_{1}\\ -\ln \left (y -1\right )+\frac {\ln \left (y \right )}{2}+\frac {\ln \left (y -2\right )}{2}&=t +c_{1} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}}{{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}-\sqrt {-{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}+1}-1}\\ &=\frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}-\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1}\\ y_2&=\frac {{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}}{{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}+\sqrt {-{\mathrm e}^{2 c_{1}} {\mathrm e}^{2 t}+1}-1}\\ &=\frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}+\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}-\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1} \\ \tag{2} y &= \frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}+\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}-\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1} \] Verified OK.
\[ y = \frac {c_{1}^{2} {\mathrm e}^{2 t}}{c_{1}^{2} {\mathrm e}^{2 t}+\sqrt {-c_{1}^{2} {\mathrm e}^{2 t}+1}-1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y \left (-2+y\right ) \left (-1+y\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y \left (-2+y\right ) \left (-1+y\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (-2+y\right ) \left (-1+y\right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y \left (-2+y\right ) \left (-1+y\right )}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (-2+y\right )}{2}-\ln \left (-1+y\right )+\frac {\ln \left (y\right )}{2}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {-\left ({\mathrm e}^{c_{1}}\right )^{2} \left ({\mathrm e}^{t}\right )^{2}+1}-1}{\sqrt {-\left ({\mathrm e}^{c_{1}}\right )^{2} \left ({\mathrm e}^{t}\right )^{2}+1}}, y=\frac {\sqrt {-\left ({\mathrm e}^{c_{1}}\right )^{2} \left ({\mathrm e}^{t}\right )^{2}+1}+1}{\sqrt {-\left ({\mathrm e}^{c_{1}}\right )^{2} \left ({\mathrm e}^{t}\right )^{2}+1}}\right \} \end {array} \]
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: 0.125 (sec). Leaf size: 75
dsolve(diff(y(t),t) = y(t)*(-2+y(t))*(-1+y(t)),y(t), singsol=all)
\begin{align*} y \left (t \right ) &= \frac {{\mathrm e}^{2 t} c_{1}}{\left (-1-\sqrt {-c_{1} {\mathrm e}^{2 t}+1}\right ) \sqrt {-c_{1} {\mathrm e}^{2 t}+1}} \\ y \left (t \right ) &= \frac {{\mathrm e}^{2 t} c_{1}}{\left (1-\sqrt {-c_{1} {\mathrm e}^{2 t}+1}\right ) \sqrt {-c_{1} {\mathrm e}^{2 t}+1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 11.055 (sec). Leaf size: 100
DSolve[y'[t] == y[t]*(-2+y[t])*(-1+y[t]),y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to \frac {-\sqrt {1+e^{2 (t+c_1)}}+e^{2 (t+c_1)}+1}{1+e^{2 (t+c_1)}} \\ y(t)\to \frac {\sqrt {1+e^{2 (t+c_1)}}+e^{2 (t+c_1)}+1}{1+e^{2 (t+c_1)}} \\ y(t)\to 0 \\ y(t)\to 1 \\ y(t)\to 2 \\ \end{align*}