Internal problem ID [6681]
Internal file name [OUTPUT/5929_Sunday_June_05_2022_04_01_59_PM_39823723/index.tex
]
Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL,
WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th
edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS.
Page 297
Problem number: 64.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "linear", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_linear, `class A`]]
\[ \boxed {y^{\prime }+y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
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) &=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime }+y = \left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \end {align*}
The domain of \(p(t)=1\) is \[
\{-\infty
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode
and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )+Y \left (s \right ) = \frac {1-2 \,{\mathrm e}^{-s}}{s}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} s Y \left (s \right )+Y \left (s \right ) = \frac {1-2 \,{\mathrm e}^{-s}}{s} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {-1+2 \,{\mathrm e}^{-s}}{s \left (s +1\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {-1+2 \,{\mathrm e}^{-s}}{s \left (s +1\right )}\right )\\ &= 1-{\mathrm e}^{-t}-2 \operatorname {Heaviside}\left (t -1\right ) \left (1-{\mathrm e}^{-t +1}\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 1-{\mathrm e}^{-t} & t <1 \\ -1-{\mathrm e}^{-t}+2 \,{\mathrm e}^{-t +1} & 1\le t \end {array}\right .
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 1-{\mathrm e}^{-t} & t <1 \\ -1-{\mathrm e}^{-t}+2 \,{\mathrm e}^{-t +1} & 1\le t \end {array}\right . \\
\end{align*} Verification of solutions
\[
y = \left \{\begin {array}{cc} 1-{\mathrm e}^{-t} & t <1 \\ -1-{\mathrm e}^{-t}+2 \,{\mathrm e}^{-t +1} & 1\le t \end {array}\right .
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \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 {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y+\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+y\right )=\mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=\mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{t} \\ {} & {} & y=\frac {\int {\mathrm e}^{t} \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ -1 & 1\le t \end {array}\right .\right )d t +c_{1}}{{\mathrm e}^{t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left \{\begin {array}{cc} 0 & t \le 0 \\ {\mathrm e}^{t}-1 & t \le 1 \\ -{\mathrm e}^{t}-1+2 \,{\mathrm e} & 1 Maple trace
✓ Solution by Maple
Time used: 2.782 (sec). Leaf size: 35
\[
y \left (t \right ) = -{\mathrm e}^{-t}-\left (\left \{\begin {array}{cc} -1 & t <1 \\ 1-2 \,{\mathrm e}^{1-t} & 1\le t \end {array}\right .\right )
\]
✓ Solution by Mathematica
Time used: 0.065 (sec). Leaf size: 43
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & t\leq 0 \\ 1-e^{-t} & 06.14.2 Solving as laplace ode
6.14.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([diff(y(t),t)+y(t)=piecewise(0<=t and t<1,1,t>=1,-1),y(0) = 0],y(t), singsol=all)
DSolve[{y'[t]+y[t]==Piecewise[{{1,0<=t<1},{-1,t>=1}}],{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]