Internal problem ID [2869]
Internal file name [OUTPUT/2361_Sunday_June_05_2022_03_00_39_AM_56517254/index.tex
]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for
10.7. page 704
Problem number: Problem 29.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "linear", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_linear, `class A`]]
\[ \boxed {y^{\prime }-y=4 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \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) &=4 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime }-y = 4 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\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 {4 \,{\mathrm e}^{-\frac {s \pi }{4}} s}{s^{2}+1}\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} s Y \left (s \right )-1-Y \left (s \right ) = \frac {4 \,{\mathrm e}^{-\frac {s \pi }{4}} s}{s^{2}+1} \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {4 \,{\mathrm e}^{-\frac {s \pi }{4}} s +s^{2}+1}{\left (s^{2}+1\right ) \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 {4 \,{\mathrm e}^{-\frac {s \pi }{4}} s +s^{2}+1}{\left (s^{2}+1\right ) \left (s -1\right )}\right )\\ &= -2 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t \right ) \sqrt {2}+{\mathrm e}^{t}+2 \left (1-\operatorname {Heaviside}\left (-t +\frac {\pi }{4}\right )\right ) {\mathrm e}^{t -\frac {\pi }{4}} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} {\mathrm e}^{t} & t <\frac {\pi }{4} \\ {\mathrm e}^{\frac {\pi }{4}}-2 & t =\frac {\pi }{4} \\ {\mathrm e}^{t}-2 \cos \left (t \right ) \sqrt {2}+2 \,{\mathrm e}^{t -\frac {\pi }{4}} & \frac {\pi }{4} The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} {\mathrm e}^{t} & t <\frac {\pi }{4} \\ {\mathrm e}^{\frac {\pi }{4}}-2 & t &=\frac {\pi }{4} \\ {\mathrm e}^{t}-2 \cos \left (t \right ) \sqrt {2}+2 \,{\mathrm e}^{t -\frac {\pi }{4}} & \frac {\pi }{4} Verification of solutions
\[
y = \left \{\begin {array}{cc} {\mathrm e}^{t} & t <\frac {\pi }{4} \\ {\mathrm e}^{\frac {\pi }{4}}-2 & t =\frac {\pi }{4} \\ {\mathrm e}^{t}-2 \cos \left (t \right ) \sqrt {2}+2 \,{\mathrm e}^{t -\frac {\pi }{4}} & \frac {\pi }{4}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y=4 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right ), 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 }=y+4 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\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=4 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\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 )=4 \mu \left (t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\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 4 \mu \left (t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int 4 \mu \left (t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 4 \mu \left (t \right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-t} \\ {} & {} & y=\frac {\int 4 \,{\mathrm e}^{-t} \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \sin \left (t +\frac {\pi }{4}\right )d t +c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {4 \left (-\frac {{\mathrm e}^{-t} \cos \left (t +\frac {\pi }{4}\right )}{2}-\frac {{\mathrm e}^{-t} \sin \left (t +\frac {\pi }{4}\right )}{2}\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )+2 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-\frac {\pi }{4}}+c_{1}}{{\mathrm e}^{-t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\left (-2 \cos \left (t +\frac {\pi }{4}\right )+2 \,{\mathrm e}^{t -\frac {\pi }{4}}-2 \sin \left (t +\frac {\pi }{4}\right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )+c_{1} {\mathrm e}^{t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (-2 \cos \left (t +\frac {\pi }{4}\right )+2 \,{\mathrm e}^{t -\frac {\pi }{4}}-2 \sin \left (t +\frac {\pi }{4}\right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )+{\mathrm e}^{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-2 \cos \left (t +\frac {\pi }{4}\right )+2 \,{\mathrm e}^{t -\frac {\pi }{4}}-2 \sin \left (t +\frac {\pi }{4}\right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )+{\mathrm e}^{t} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 2.329 (sec). Leaf size: 40
\[
y \left (t \right ) = \left (-2 \cos \left (t \right ) \sqrt {2}+2 \,{\mathrm e}^{t -\frac {\pi }{4}}\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right )+{\mathrm e}^{t}
\]
✓ Solution by Mathematica
Time used: 0.112 (sec). Leaf size: 40
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^t & 4 t\leq \pi \\ -2 \sqrt {2} \cos (t)+e^t+2 e^{t-\frac {\pi }{4}} & \text {True} \\ \end {array} \\ \end {array}
\]
14.3.2 Solving as laplace ode
14.3.3 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)=4*Heaviside(t-Pi/4)*cos(t-Pi/4),y(0) = 1],y(t), singsol=all)
DSolve[{y'[t]-y[t]==4*UnitStep[t-Pi/4]*Cos[t-Pi/4],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]