3.12 problem 16

Internal problem ID [12939]
Internal file name [OUTPUT/11592_Tuesday_November_07_2023_11_27_35_PM_70610284/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 1.4 page 61
Problem number: 16.
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]

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

3.12.1 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) &=1\\ q(t) &=2 \end {align*}

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

The domain of \(p(t)=1\) is \[ \{-\infty

3.12.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(t=0\) and \(y=1\) 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 ) = -i \pi +t \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 ) &= -i \pi +t \\ \end{align*}

Verification of solutions

\[ -\ln \left (y-2\right ) = -i \pi +t \] Verified OK.

3.12.3 Maple step by step solution

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

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),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = 2-{\mathrm e}^{-t} \]

✓ Solution by Mathematica

Time used: 0.039 (sec). Leaf size: 14

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

\[ y(t)\to 2-e^{-t} \]