Internal problem ID [13285]
Internal file name [OUTPUT/12457_Wednesday_February_14_2024_02_06_21_AM_92802594/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 b.
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]
\[ \boxed {y^{\prime }=\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 2] \end {align*}
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 \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime } = \left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x \end {array}\right . \end {align*}
The domain of \(p(x)=0\) is \[
\{-\infty
Integrating both sides gives \begin {align*} y &= \int { \left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x \end {array}\right .\,\mathop {\mathrm {d}x}}\\ &= \left (\left \{\begin {array}{cc} 0 & x \le 1 \\ x -1 & 1 Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=2\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 2 = c_{1} \end {align*}
The solutions are \begin {align*} c_{1} = 2 \end {align*}
Trying the constant \begin {align*} c_{1} = 2 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\left \{\begin {array}{cc} 2 & x \le 1 \\ 1+x & 1 But this does not satisfy the initial conditions. Hence no solution can be found. The constant
\(c_{1} = 2\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place.
The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 2 & x \le 1 \\ 1+x & 1 Verification of solutions
\[
y = \left \{\begin {array}{cc} 2 & x \le 1 \\ 1+x & 1
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\left \{\begin {array}{cc} 0 & x <1 \\ 1 & 1\le x \end {array}\right ., y \left (0\right )=2\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 \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 & 1 Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 15
\[
y \left (x \right ) = \left \{\begin {array}{cc} 2 & x <1 \\ 1+x & 1\le x \end {array}\right .
\]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 15
\[
y(x)\to \begin {array}{cc} \{ & \begin {array}{cc} 2 & x\leq 1 \\ x+1 & \text {True} \\ \end {array} \\ \end {array}
\]
1.44.2 Solving as quadrature ode
1.44.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
dsolve([diff(y(x),x)=piecewise(x<1,0,x>=1,1),y(0) = 2],y(x), singsol=all)
DSolve[{y'[x]==Piecewise[{{0,x<1},{1,x>=1}}],{y[0]==2}},y[x],x,IncludeSingularSolutions -> True]