4.1 problem 1

Internal problem ID [54]
Internal file name [OUTPUT/54_Sunday_June_05_2022_01_34_08_AM_49603279/index.tex]

Book: Differential equations and linear algebra, 3rd ed., Edwards and Penney
Section: Section 1.5. Linear first order equations. Page 56
Problem number: 1.
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 ) = 0] \end {align*}

4.1.1 Existence and uniqueness analysis

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

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

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

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

4.1.2 Solving as quadrature ode

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\ln \left (2\right )-i \pi = c_{1} \end {align*}

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

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

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\ln \left (y -2\right ) = x -\ln \left (2\right )-i \pi \end {align*}

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

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\ln \left (y-2\right ) &= x -\ln \left (2\right )-i \pi \\ \end{align*}

Verification of solutions

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

4.1.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y=2, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y+2 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-y+2}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-y+2}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (-y+2\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-{\mathrm e}^{-x -c_{1}}+2 \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-{\mathrm e}^{-c_{1}}+2 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =-\ln \left (2\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =-\ln \left (2\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-2 \,{\mathrm e}^{-x}+2 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-2 \,{\mathrm e}^{-x}+2 \end {array} \]

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: 0.015 (sec). Leaf size: 12

dsolve([y(x)+diff(y(x),x) = 2,y(0) = 0],y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.021 (sec). Leaf size: 14

DSolve[{y[x]+y'[x] == 2,y[0]==0},y[x],x,IncludeSingularSolutions -> True]
 

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