Internal problem ID [843]
Internal file name [OUTPUT/843_Sunday_June_05_2022_01_50_58_AM_84931335/index.tex
]
Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima,
Meade
Section: Chapter 6.2, The Laplace Transform. Solution of Initial Value Problems. page
255
Problem number: 18.
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]]
\[ \boxed {y^{\prime \prime }+4 y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t <\infty \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0] \end {align*}
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) &=4\\ F &=\left \{\begin {array}{cc} 0 & t <0 \\ 1 & t <1 \\ 0 & 1\le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+4 y = \left \{\begin {array}{cc} 0 & t <0 \\ 1 & 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 )+4 Y \left (s \right ) = \frac {1-{\mathrm e}^{-s}}{s}\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 )+4 Y \left (s \right ) = \frac {1-{\mathrm e}^{-s}}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {-1+{\mathrm e}^{-s}}{s \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 {-1+{\mathrm e}^{-s}}{s \left (s^{2}+4\right )}\right )\\ &= -\frac {\cos \left (2 t \right )}{4}+\frac {1}{4}-\frac {\operatorname {Heaviside}\left (-1+t \right ) \sin \left (-1+t \right )^{2}}{2} \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -\frac {\cos \left (2 t \right )}{4}+\frac {1}{4} & t <1 \\ -\frac {\cos \left (2 t \right )}{4}+\frac {1}{4}-\frac {\sin \left (-1+t \right )^{2}}{2} & 1\le t \end {array}\right .
\] Simplifying the solution gives \[
y = -\frac {\cos \left (2 t \right )}{4}+\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cos \left (2 t -2\right ) & 1\le t \end {array}\right .\right )}{4}
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= -\frac {\cos \left (2 t \right )}{4}+\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cos \left (2 t -2\right ) & 1\le t \end {array}\right .\right )}{4} \\
\end{align*} Verification of solutions
\[
y = -\frac {\cos \left (2 t \right )}{4}+\frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cos \left (2 t -2\right ) & 1\le t \end {array}\right .\right )}{4}
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y=\left \{\begin {array}{cc} 0 & t <0 \\ 1 & 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 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2 \,\mathrm {I}, 2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (2 t \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (2 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 (2 t \right )+c_{2} \sin \left (2 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 <0 \\ 1 & 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 (2 t \right ) & \sin \left (2 t \right ) \\ -2 \sin \left (2 t \right ) & 2 \cos \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\cos \left (2 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ \frac {\sin \left (2 t \right )}{2} & t <1 \\ 0 & 1\le t \end {array}\right .\right )d t \right )+\sin \left (2 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t <0 \\ \frac {\cos \left (2 t \right )}{2} & t <1 \\ 0 & 1\le t \end {array}\right .\right )d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\left (\left \{\begin {array}{cc} 0 & t \le 0 \\ -1+\cos \left (2 t \right ) & t \le 1 \\ 2 \sin \left (1\right )^{2} \cos \left (2 t \right )-\sin \left (2 t \right ) \sin \left (2\right ) & 1 Maple trace
✓ Solution by Maple
Time used: 0.875 (sec). Leaf size: 35
\[
y \left (t \right ) = \frac {\left (\left \{\begin {array}{cc} 1 & t <1 \\ \cos \left (2 t -2\right ) & 1\le t \end {array}\right .\right )}{4}-\frac {\cos \left (2 t \right )}{4}
\]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 39
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & t\leq 0 \\ \frac {\sin ^2(t)}{2} & 03.11.2 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)+4*y(t)=piecewise(0<=t and t<1,1,1<=t and t<infinity,0),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+4*y[t]==Piecewise[{{1,0<t<1},{0,1<=t<Infinity}}],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]