Internal problem ID [5698]
Internal file name [OUTPUT/4946_Sunday_June_05_2022_03_11_18_PM_59599965/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: 22.
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 }+3 y^{\prime }+2 y=\left \{\begin {array}{cc} 4 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) &=3\\ q(t) &=2\\ F &=4 \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & t =1 \\ 2 & 1 Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }+2 y = 4 \left (\left \{\begin {array}{cc} 0 & t \le 0 \\ t & t <1 \\ 0 & t =1 \\ 2 & 1 The domain of \(p(t)=3\) 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 )+3 s Y \left (s \right )-3 y \left (0\right )+2 Y \left (s \right ) = \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 )+3 s Y \left (s \right )+2 Y \left (s \right ) = \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {\mathcal {L}\left (\left \{\begin {array}{cc} 0 & t \le 0 \\ 4 t & t <1 \\ 0 & t =1 \\ 8 & 1 Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {\mathcal {L}\left (\left \{\begin {array}{cc} 0 & t \le 0 \\ 4 t & t <1 \\ 0 & t =1 \\ 8 & 1 Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} -3+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+2 t & t <1 \\ -6+4 \,{\mathrm e}^{-1}-{\mathrm e}^{-2} & t =1 \\ 4+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{-2 t +2}-8 \,{\mathrm e}^{1-t} & 1 The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} -3+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+2 t & t <1 \\ -6+4 \,{\mathrm e}^{-1}-{\mathrm e}^{-2} & t &=1 \\ 4+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{-2 t +2}-8 \,{\mathrm e}^{1-t} & 1 Verification of solutions
\[
y = \left \{\begin {array}{cc} -3+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+2 t & t <1 \\ -6+4 \,{\mathrm e}^{-1}-{\mathrm e}^{-2} & t =1 \\ 4+4 \,{\mathrm e}^{-t}-{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{-2 t +2}-8 \,{\mathrm e}^{1-t} & 1
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }+3 y^{\prime }+2 y=\left \{\begin {array}{cc} 0 & t \le 0 \\ 4 t & t <1 \\ 0 & t =1 \\ 8 & 1 Maple trace
✓ Solution by Maple
Time used: 1.156 (sec). Leaf size: 71
\[
y \left (t \right ) = \left \{\begin {array}{cc} 2 t -{\mathrm e}^{-2 t}-3+4 \,{\mathrm e}^{-t} & t <1 \\ -{\mathrm e}^{-2}+1+4 \,{\mathrm e}^{-1} & t =1 \\ 3 \,{\mathrm e}^{-2 t +2}-8 \,{\mathrm e}^{1-t}-{\mathrm e}^{-2 t}+4+4 \,{\mathrm e}^{-t} & 1 ✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 70
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & t\leq 0 \\ 2 t-e^{-2 t}+4 e^{-t}-3 & 07.5.1 Existence and uniqueness analysis
7.5.2 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)]
<- double symmetry of the form [xi=0, eta=F(x)] successful`
dsolve([diff(y(t),t$2)+3*diff(y(t),t)+2*y(t)=piecewise(0<t and t<1,4*t,t>1,8),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
DSolve[{y''[t]+3*y'[t]+2*y[t]==Piecewise[{{4*t,0<t<1},{8,t>1}}],{y[0]==0,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]