Internal problem ID [5699]
Internal file name [OUTPUT/4947_Sunday_June_05_2022_03_11_22_PM_35045146/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number: 23.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[
\boxed {y^{\prime \prime }+y^{\prime }-2 y=\left \{\begin {array}{cc} 3 \sin \left (t \right )-\cos \left (t \right ) & 0
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}
Where here \begin {align*} p(t) &=1\\ q(t) &=-2\\ F &=\left \{\begin {array}{cc} 0 & t \le 0 \\ 3 \sin \left (t \right )-\cos \left (t \right ) & t <2 \pi \\ 0 & t =2 \pi \\ 3 \sin \left (2 t \right )-\cos \left (2 t \right ) & 2 \pi Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime }-2 y = \left \{\begin {array}{cc} 0 & t \le 0 \\ 3 \sin \left (t \right )-\cos \left (t \right ) & t <2 \pi \\ 0 & t =2 \pi \\ 3 \sin \left (2 t \right )-\cos \left (2 t \right ) & 2 \pi The domain of \(p(t)=1\) is \[
\{-\infty Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (y\right ) =Y(s) \end {align*}
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 )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+s Y \left (s \right )-y \left (0\right )-2 Y \left (s \right ) = \frac {3-s +\frac {3 \,{\mathrm e}^{-2 \pi s} \left (s +2\right ) \left (s -1\right )}{s^{2}+4}}{s^{2}+1}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-1-s +s Y \left (s \right )-2 Y \left (s \right ) = \frac {3-s +\frac {3 \,{\mathrm e}^{-2 \pi s} \left (s +2\right ) \left (s -1\right )}{s^{2}+4}}{s^{2}+1} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{4}-s^{3}+3 \,{\mathrm e}^{-2 \pi s} s +6 s^{2}-3 \,{\mathrm e}^{-2 \pi s}-4 s +8}{\left (s -1\right ) \left (s^{2}+1\right ) \left (s^{2}+4\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {s^{4}-s^{3}+3 \,{\mathrm e}^{-2 \pi s} s +6 s^{2}-3 \,{\mathrm e}^{-2 \pi s}-4 s +8}{\left (s -1\right ) \left (s^{2}+1\right ) \left (s^{2}+4\right )}\right )\\ &= -\sin \left (t \right )+{\mathrm e}^{t}+\frac {\operatorname {Heaviside}\left (t -2 \pi \right ) \left (-\sin \left (2 t \right )+2 \sin \left (t \right )\right )}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -\sin \left (t \right )+{\mathrm e}^{t} & t <2 \pi \\ {\mathrm e}^{t}-\frac {\sin \left (2 t \right )}{2} & 2 \pi \le t \end {array}\right .
\] Simplifying the solution gives \[
y = {\mathrm e}^{t}-\left (\left \{\begin {array}{cc} \sin \left (t \right ) & t <2 \pi \\ \frac {\sin \left (2 t \right )}{2} & 2 \pi \le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= {\mathrm e}^{t}-\left (\left \{\begin {array}{cc} \sin \left (t \right ) & t <2 \pi \\ \frac {\sin \left (2 t \right )}{2} & 2 \pi \le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = {\mathrm e}^{t}-\left (\left \{\begin {array}{cc} \sin \left (t \right ) & t <2 \pi \\ \frac {\sin \left (2 t \right )}{2} & 2 \pi \le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+y^{\prime }-2 y=\left \{\begin {array}{cc} 0 & t \le 0 \\ 3 \sin \left (t \right )-\cos \left (t \right ) & t <2 \pi \\ 0 & t =2 \pi \\ 3 \sin \left (2 t \right )-\cos \left (2 t \right ) & 2 \pi Maple trace
✓ Solution by Maple
Time used: 1.344 (sec). Leaf size: 31
\[
y \left (t \right ) = {\mathrm e}^{t}-\left (\left \{\begin {array}{cc} \sin \left (t \right ) & t <2 \pi \\ \frac {\sin \left (2 t \right )}{2} & 2 \pi \le t \end {array}\right .\right )
\]
✓ Solution by Mathematica
Time used: 0.058 (sec). Leaf size: 55
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {e^{-2 t}}{3}+\frac {2 e^t}{3} & t\leq 0 \\ e^t-\sin (t) & 07.6.1 Existence and uniqueness analysis
7.6.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful`
dsolve([diff(y(t),t$2)+diff(y(t),t)-2*y(t)=piecewise(0<t and t<2*Pi,3*sin(t)-cos(t),t>2*Pi,3*sin(2*t)-cos(2*t)),y(0) = 1, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+y'[t]-2*y[t]==Piecewise[{{3*Sin[t]-Cos[t],0<t<2*Pi},{3*Sin[2*t]-Cos[2*t],t>2*Pi}}],{y[0]==1,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]