1.45 problem 2.9 c

Internal problem ID [13286]
Internal file name [OUTPUT/12458_Wednesday_February_14_2024_02_06_21_AM_39720513/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 2. Integration and differential equations. Additional exercises. page 32
Problem number: 2.9 c.
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]

\[ \boxed {y^{\prime }=\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}

1.45.1 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) &=0\\ q(x) &=\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime } = \left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right . \end {align*}

The domain of \(p(x)=0\) is \[ \{-\infty

1.45.2 Solving as quadrature ode

Integrating both sides gives \begin {align*} y &= \int { \left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right .\,\mathop {\mathrm {d}x}}\\ &= \left (\left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & x \le 2 \\ 1 & 2

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*} 0 = c_{1} \end {align*}

The solutions are \begin {align*} c_{1} = 0 \end {align*}

Trying the constant \begin {align*} c_{1} = 0 \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & x \le 2 \\ 1 & 2

But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{1} = 0\) does not give valid solution.

Which is valid for any constant of integration. Therefore keeping the constant in place.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & x \le 2 \\ 1 & 2

Verification of solutions

\[ y = \left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & x \le 2 \\ 1 & 2

1.45.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right ., y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x <2 \\ 0 & 2\le x \end {array}\right .\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & x \le 2 \\ 1 & 2

Maple trace

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

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 19

dsolve([diff(y(x),x)=piecewise(x<1,0,1<=x and x<2,1,2<=x,0),y(0) = 0],y(x), singsol=all)
 

\[ y \left (x \right ) = \left \{\begin {array}{cc} 0 & x <1 \\ -1+x & x <2 \\ 1 & 2\le x \end {array}\right . \]

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 23

DSolve[{y'[x]==Piecewise[{{0,x<1},{1,1<=x<2},{0,2<=x}}],{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & x\leq 1 \\ x-1 & 1