1.28 problem 31

Internal problem ID [12891]
Internal file name [OUTPUT/11544_Monday_November_06_2023_01_33_12_PM_92578670/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 1.2. page 33
Problem number: 31.
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]

\[ \boxed {y^{\prime }-2 y=1} \] With initial conditions \begin {align*} [y \left (0\right ) = 3] \end {align*}

1.28.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=-2\\ q(t) &=1 \end {align*}

Hence the ode is \begin {align*} y^{\prime }-2 y = 1 \end {align*}

The domain of \(p(t)=-2\) is \[ \{-\infty

1.28.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} \int \frac {1}{2 y +1}d y &= \int {dt}\\ \frac {\ln \left (y +\frac {1}{2}\right )}{2}&= 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 (7\right )}{2}-\frac {\ln \left (2\right )}{2} = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {\ln \left (y +\frac {1}{2}\right )}{2} = t +\frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2} \end {align*}

The constant \(c_{1} = \frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (2 y+1\right )}{2} &= t +\frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2} \\ \end{align*}

Verification of solutions

\[ -\frac {\ln \left (2\right )}{2}+\frac {\ln \left (2 y+1\right )}{2} = t +\frac {\ln \left (7\right )}{2}-\frac {\ln \left (2\right )}{2} \] Verified OK.

1.28.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=1, y \left (0\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{2 y+1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{2 y+1}d t =\int 1d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (2 y+1\right )}{2}=t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{2 t +2 c_{1}}}{2}-\frac {1}{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=\frac {{\mathrm e}^{2 c_{1}}}{2}-\frac {1}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (7\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (7\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {7 \,{\mathrm e}^{2 t}}{2}-\frac {1}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {7 \,{\mathrm e}^{2 t}}{2}-\frac {1}{2} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 12

dsolve([diff(y(t),t)=2*y(t)+1,y(0) = 3],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {1}{2}+\frac {7 \,{\mathrm e}^{2 t}}{2} \]

✓ Solution by Mathematica

Time used: 0.045 (sec). Leaf size: 18

DSolve[{y'[t]==2*y[t]+1,{y[0]==3}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{2} \left (7 e^{2 t}-1\right ) \]