Internal problem ID [5701]
Internal file name [OUTPUT/4949_Sunday_June_05_2022_03_11_32_PM_75529370/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: 25.
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=\left \{\begin {array}{cc} t & 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) &=0\\ q(t) &=1\\ F &=\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & 1\le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+y = \left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & 1\le t \end {array}\right . \end {align*}
The domain of \(p(t)=0\) 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 )+Y \left (s \right ) = \frac {-\left (s +1\right ) {\mathrm e}^{-s}+1}{s^{2}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=0 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+Y \left (s \right ) = \frac {-\left (s +1\right ) {\mathrm e}^{-s}+1}{s^{2}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {{\mathrm e}^{-s} s +{\mathrm e}^{-s}-1}{s^{2} \left (s^{2}+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 {{\mathrm e}^{-s} s +{\mathrm e}^{-s}-1}{s^{2} \left (s^{2}+1\right )}\right )\\ &= -\sin \left (t \right )+t -\operatorname {Heaviside}\left (t -1\right ) \left (2 \sin \left (\frac {t}{2}-\frac {1}{2}\right )^{2}-\sin \left (t -1\right )+t -1\right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -\sin \left (t \right )+t & t <1 \\ -\sin \left (t \right )-2 \sin \left (\frac {t}{2}-\frac {1}{2}\right )^{2}+\sin \left (t -1\right )+1 & 1\le t \end {array}\right .
\] Simplifying the solution gives \[
y = -\sin \left (t \right )+\left (\left \{\begin {array}{cc} t & t <1 \\ \cos \left (t -1\right )+\sin \left (t -1\right ) & 1\le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\sin \left (t \right )+\left (\left \{\begin {array}{cc} t & t <1 \\ \cos \left (t -1\right )+\sin \left (t -1\right ) & 1\le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = -\sin \left (t \right )+\left (\left \{\begin {array}{cc} t & t <1 \\ \cos \left (t -1\right )+\sin \left (t -1\right ) & 1\le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+y=\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & 1\le t \end {array}\right ., y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (t \right )+c_{2} \sin \left (t \right )+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & 1\le t \end {array}\right .\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} \cos \left (t \right ) & \sin \left (t \right ) \\ -\sin \left (t \right ) & \cos \left (t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\cos \left (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ \sin \left (t \right ) t & t <1 \\ 0 & 1\le t \end {array}\right .\right )d t \right )+\sin \left (t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ \cos \left (t \right ) t & t <1 \\ 0 & 1\le t \end {array}\right .\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\left \{\begin {array}{cc} 0 & t \le 0 \\ -\sin \left (t \right )+t & t \le 1 \\ \left (-\sin \left (1\right )+\cos \left (1\right )\right ) \cos \left (t \right )+\left (-1+\sin \left (1\right )+\cos \left (1\right )\right ) \sin \left (t \right ) & 1 Maple trace
✓ Solution by Maple
Time used: 1.61 (sec). Leaf size: 37
\[
y \left (t \right ) = -\sin \left (t \right )+\left (\left \{\begin {array}{cc} t & t <1 \\ \sin \left (t -1\right )+\cos \left (t -1\right ) & 1\le t \end {array}\right .\right )
\]
✓ Solution by Mathematica
Time used: 0.034 (sec). Leaf size: 44
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} t-\sin (t) & 07.8.1 Existence and uniqueness analysis
7.8.2 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)+y(t)=piecewise(0<t and t<1,t,t>1,0),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+y[t]==Piecewise[{{t,0<t<1},{0,t>1}}],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]