Internal problem ID [5703]
Internal file name [OUTPUT/4951_Sunday_June_05_2022_03_14_35_PM_63270777/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: 27.
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} 8 t^{2} & 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) &=4\\ F &=\left \{\begin {array}{cc} 0 & t \le 0 \\ 8 t^{2} & t <5 \\ 0 & 5\le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+4 y = \left \{\begin {array}{cc} 0 & t \le 0 \\ 8 t^{2} & t <5 \\ 0 & 5\le t \end {array}\right . \end {align*}
The domain of \(p(t)=0\) is \[
\{-\infty Since both initial conditions are not at zero, then let \begin {align*} y(0) &= c_{1}\\ y'(0) &= c_{2} \end {align*}
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 {16-8 \left (25 s^{2}+10 s +2\right ) {\mathrm e}^{-5 s}}{s^{3}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +4 Y \left (s \right ) = \frac {16-8 \left (25 s^{2}+10 s +2\right ) {\mathrm e}^{-5 s}}{s^{3}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {-s^{4} c_{1} -c_{2} s^{3}+200 \,{\mathrm e}^{-5 s} s^{2}+80 \,{\mathrm e}^{-5 s} s +16 \,{\mathrm e}^{-5 s}-16}{s^{3} \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} c_{1} -c_{2} s^{3}+200 \,{\mathrm e}^{-5 s} s^{2}+80 \,{\mathrm e}^{-5 s} s +16 \,{\mathrm e}^{-5 s}-16}{s^{3} \left (s^{2}+4\right )}\right )\\ &= -1+\frac {c_{2} \sin \left (2 t \right )}{2}+2 t^{2}+\cos \left (2 t \right ) \left (1+c_{1} \right )-\operatorname {Heaviside}\left (t -5\right ) \left (100 \sin \left (t -5\right )^{2}+2 t^{2}+\cos \left (-10+2 t \right )-10 \sin \left (-10+2 t \right )-51\right ) \end {align*}
Since both initial conditions given are not at zero, then we need to setup two equations to
solve for \(c_{1},c_{1}\). At \(t=1\) the first equation becomes, using the above solution \begin {align*} 1+\cos \left (2\right ) &= 1+\frac {c_{2} \sin \left (2\right )}{2}+\cos \left (2\right ) \left (1+c_{1} \right ) \end {align*}
And taking derivative of the solution and evaluating at \(t=1\) gives the second equation as
\begin {align*} 4-2 \sin \left (2\right ) &= c_{2} \cos \left (2\right )+4-2 \sin \left (2\right ) \left (1+c_{1} \right ) \end {align*}
Solving gives \begin {align*} c_{1} &= 0\\ c_{2} &= 0 \end {align*}
Subtituting these in the solution obtained above gives \begin {align*} y &= -1+2 t^{2}+\cos \left (2 t \right )-\operatorname {Heaviside}\left (t -5\right ) \left (100 \sin \left (t -5\right )^{2}+2 t^{2}+\cos \left (-10+2 t \right )-10 \sin \left (-10+2 t \right )-51\right )\\ &= -2 \operatorname {Heaviside}\left (t -5\right ) t^{2}+2 t^{2}+\operatorname {Heaviside}\left (t -5\right )-1+49 \operatorname {Heaviside}\left (t -5\right ) \cos \left (-10+2 t \right )+10 \operatorname {Heaviside}\left (t -5\right ) \sin \left (-10+2 t \right )+\cos \left (2 t \right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -1+2 t^{2}+\cos \left (2 t \right ) & t <5 \\ \cos \left (2 t \right )+49 \cos \left (-10+2 t \right )+10 \sin \left (-10+2 t \right ) & 5\le t \end {array}\right .
\] Simplifying the solution gives \[
y = \cos \left (2 t \right )+\left (\left \{\begin {array}{cc} 2 t^{2}-1 & t <5 \\ 49 \cos \left (-10+2 t \right )+10 \sin \left (-10+2 t \right ) & 5\le t \end {array}\right .\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= \cos \left (2 t \right )+\left (\left \{\begin {array}{cc} 2 t^{2}-1 & t <5 \\ 49 \cos \left (-10+2 t \right )+10 \sin \left (-10+2 t \right ) & 5\le t \end {array}\right .\right ) \\
\end{align*} Verification of solutions
\[
y = \cos \left (2 t \right )+\left (\left \{\begin {array}{cc} 2 t^{2}-1 & t <5 \\ 49 \cos \left (-10+2 t \right )+10 \sin \left (-10+2 t \right ) & 5\le t \end {array}\right .\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+4 y=\left \{\begin {array}{cc} 0 & t \le 0 \\ 8 t^{2} & t <5 \\ 0 & 5\le t \end {array}\right ., y \left (1\right )=1+\cos \left (2\right ), y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=4-2 \sin \left (2\right )\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 \le 0 \\ 8 t^{2} & t <5 \\ 0 & 5\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 \le 0 \\ 4 \sin \left (2 t \right ) t^{2} & t <5 \\ 0 & 5\le t \end {array}\right .\right )d t \right )+\sin \left (2 t \right ) \left (\int \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ 4 \cos \left (2 t \right ) t^{2} & t <5 \\ 0 & 5\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 \\ -1+2 t^{2}+\cos \left (2 t \right ) & t \le 5 \\ \left (49 \sin \left (10\right )+10 \cos \left (10\right )\right ) \sin \left (2 t \right )+49 \cos \left (2 t \right ) \left (\cos \left (10\right )-\frac {10 \sin \left (10\right )}{49}+\frac {1}{49}\right ) & 5 Maple trace
✓ Solution by Maple
Time used: 1.844 (sec). Leaf size: 87
\[
y \left (t \right ) = \cos \left (2 t \right )+\left (\left \{\begin {array}{cc} 2 t^{2}-1 & t <5 \\ 10 \sin \left (2 t -10\right )+49 \cos \left (2 t -10\right ) & 5\le t \end {array}\right .\right )
\]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 51
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 2 t^2+\cos (2 t)-1 & 07.10.1 Existence and uniqueness analysis
7.10.2 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<5,8*t^2,t>5,0),y(1) = 1+cos(2), D(y)(1) = 4-2*sin(2)],y(t), singsol=all)
DSolve[{y''[t]+4*y[t]==Piecewise[{{8*t^2,0<t<5},{0,t>5}}],{y[1]==1+Cos[2],y'[1]==4-2*Sin[2]}},y[t],t,IncludeSingularSolutions -> True]