Internal problem ID [4949]
Internal file name [OUTPUT/4442_Sunday_June_05_2022_02_56_43_PM_76292468/index.tex]
Book: Fundamentals of Differential Equations. By Nagle, Saff and Snider. 9th edition. Boston.
Pearson 2018.
Section: Chapter 2, First order differential equations. Section 2.2, Separable Equations. Exercises.
page 46
Problem number: 32.
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]
\[ \boxed {y^{\prime }-y^{2}+3 y=2} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {3}{2}}\right ] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= y^{2}-3 y +2 \end {align*}
The \(y\) domain of \(f(x,y)\) when \(x=0\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}-3 y +2}d y &= \int {dx}\\ \ln \left (y -2\right )-\ln \left (y -1\right )&= x +c_{1} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y={\frac {3}{2}}\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} i \pi = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = i \pi \end {align*}
Trying the constant \begin {align*} c_{1} = i \pi \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \ln \left (y -2\right )-\ln \left (y -1\right ) = i \pi +x \end {align*}
The constant \(c_{1} = i \pi \) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} \ln \left (y-2\right )-\ln \left (y-1\right ) &= i \pi +x \\
\end{align*} Verification of solutions
\[
\ln \left (y-2\right )-\ln \left (y-1\right ) = i \pi +x
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}+3 y=2, y \left (0\right )=\frac {3}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-3 y+2 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}-3 y+2}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}-3 y+2}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y-2\right )-\ln \left (y-1\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {-2+{\mathrm e}^{x +c_{1}}}{{\mathrm e}^{x +c_{1}}-1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {3}{2} \\ {} & {} & \frac {3}{2}=\frac {-2+{\mathrm e}^{c_{1}}}{{\mathrm e}^{c_{1}}-1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\mathrm {I} \pi \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\mathrm {I} \pi \hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {2+{\mathrm e}^{x}}{{\mathrm e}^{x}+1} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {2+{\mathrm e}^{x}}{{\mathrm e}^{x}+1} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 15
\[
y \left (x \right ) = \frac {{\mathrm e}^{x}+2}{1+{\mathrm e}^{x}}
\]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 18
\[
y(x)\to \frac {e^x+2}{e^x+1}
\]
1.38.2 Solving as quadrature ode
1.38.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful`
dsolve([diff(y(x),x)=y(x)^2-3*y(x)+2,y(0) = 3/2],y(x), singsol=all)
DSolve[{y'[x]==y[x]^2-3*y[x]+2,{y[0]==3/2}},y[x],x,IncludeSingularSolutions -> True]