Internal problem ID [12333]
Internal file name [OUTPUT/10986_Monday_October_02_2023_02_47_43_AM_35630610/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page
368
Problem number: Problem 4(b).
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]]
\[ \boxed {y^{\prime \prime }-3 y^{\prime }+2 y=\left \{\begin {array}{cc} 0 & 0\le t <1 \\ 1 & 1\le t <2 \\ -1 & 2\le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = -1] \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) &=-3\\ q(t) &=2\\ F &=\left \{\begin {array}{cc} 0 & t <1 \\ 1 & t <2 \\ -1 & 2\le t \end {array}\right . \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-3 y^{\prime }+2 y = \left \{\begin {array}{cc} 0 & t <1 \\ 1 & t <2 \\ -1 & 2\le t \end {array}\right . \end {align*}
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 ) = \frac {{\mathrm e}^{-s}-2 \,{\mathrm e}^{-2 s}}{s}\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=-1 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+10-3 s -3 s Y \left (s \right )+2 Y \left (s \right ) = \frac {{\mathrm e}^{-s}-2 \,{\mathrm e}^{-2 s}}{s} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{2}+{\mathrm e}^{-s}-2 \,{\mathrm e}^{-2 s}-10 s}{s \left (s^{2}-3 s +2\right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3 s^{2}+{\mathrm e}^{-s}-2 \,{\mathrm e}^{-2 s}-10 s}{s \left (s^{2}-3 s +2\right )}\right )\\ &= \frac {\operatorname {Heaviside}\left (t -1\right )}{2}-\operatorname {Heaviside}\left (t -2\right )+7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}-{\mathrm e}^{2 t -4}+\frac {\left (-{\mathrm e}^{-2+2 t}+2 \,{\mathrm e}^{t -1}\right ) \operatorname {Heaviside}\left (1-t \right )}{2}+\left (-2 \,{\mathrm e}^{t -2}+{\mathrm e}^{2 t -4}\right ) \operatorname {Heaviside}\left (2-t \right ) \end {align*}
Converting the above solution to piecewise it becomes \[
y = \left \{\begin {array}{cc} 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t} & t <1 \\ 7 \,{\mathrm e}-4 \,{\mathrm e}^{2}+\frac {1}{2} & t =1 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}+\frac {1}{2} & t <2 \\ \frac {15 \,{\mathrm e}^{2}}{2}-4 \,{\mathrm e}^{4}-{\mathrm e}-\frac {1}{2} & t =2 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}-{\mathrm e}^{2 t -4}-\frac {1}{2} & 2 The solution(s) found are the following \begin{align*}
\tag{1} y &= \left \{\begin {array}{cc} 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t} & t <1 \\ 7 \,{\mathrm e}-4 \,{\mathrm e}^{2}+\frac {1}{2} & t &=1 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}+\frac {1}{2} & t <2 \\ \frac {15 \,{\mathrm e}^{2}}{2}-4 \,{\mathrm e}^{4}-{\mathrm e}-\frac {1}{2} & t &=2 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}-{\mathrm e}^{2 t -4}-\frac {1}{2} & 2 Verification of solutions
\[
y = \left \{\begin {array}{cc} 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t} & t <1 \\ 7 \,{\mathrm e}-4 \,{\mathrm e}^{2}+\frac {1}{2} & t =1 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}+\frac {1}{2} & t <2 \\ \frac {15 \,{\mathrm e}^{2}}{2}-4 \,{\mathrm e}^{4}-{\mathrm e}-\frac {1}{2} & t =2 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}+\frac {{\mathrm e}^{-2+2 t}}{2}-{\mathrm e}^{t -1}-{\mathrm e}^{2 t -4}-\frac {1}{2} & 2
\[ \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 <1 \\ 1 & t <2 \\ -1 & 2\le t \end {array}\right ., y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=-1\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}-3 r +2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 t} \\ \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} {\mathrm e}^{t}+c_{2} {\mathrm e}^{2 t}+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 <1 \\ 1 & t <2 \\ -1 & 2\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} {\mathrm e}^{t} & {\mathrm e}^{2 t} \\ {\mathrm e}^{t} & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-{\mathrm e}^{t} \left (\int \left (\left \{\begin {array}{cc} 0 & t <1 \\ 1 & t <2 \\ -1 & 2\le t \end {array}\right .\right ) {\mathrm e}^{-t}d t \right )+{\mathrm e}^{2 t} \left (\int \left (\left \{\begin {array}{cc} 0 & t <1 \\ 1 & t <2 \\ -1 & 2\le t \end {array}\right .\right ) {\mathrm e}^{-2 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\left (\left \{\begin {array}{cc} 0 & t \le 1 \\ 2 \,{\mathrm e}^{t -1}-1-{\mathrm e}^{-2+2 t} & t \le 2 \\ 2 \,{\mathrm e}^{t -1}-4 \,{\mathrm e}^{t -2}+1-{\mathrm e}^{-2+2 t}+2 \,{\mathrm e}^{2 t -4} & 2 Maple trace
✓ Solution by Maple
Time used: 7.172 (sec). Leaf size: 121
\[
y \left (t \right ) = \left \{\begin {array}{cc} 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t} & t <1 \\ 7 \,{\mathrm e}-4 \,{\mathrm e}^{2}+\frac {1}{2} & t =1 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}-{\mathrm e}^{t -1}+\frac {{\mathrm e}^{2 t -2}}{2}+\frac {1}{2} & t <2 \\ \frac {15 \,{\mathrm e}^{2}}{2}-4 \,{\mathrm e}^{4}-{\mathrm e}-\frac {1}{2} & t =2 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}-{\mathrm e}^{t -1}-{\mathrm e}^{2 t -4}+\frac {{\mathrm e}^{2 t -2}}{2}-\frac {1}{2} & 2 ✓ Solution by Mathematica
Time used: 0.068 (sec). Leaf size: 109
\[
y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^t \left (7-4 e^t\right ) & t\leq 1 \\ \frac {1}{2} \left (1-2 e^{t-1}+14 e^t-8 e^{2 t}+e^{2 t-2}\right ) & 14.26.2 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<1,0,t>=1 and t<2,1,t>=2,-1 ),y(0) = 3, D(y)(0) = -1],y(t), singsol=all)
DSolve[{y''[t]-3*y'[t]+2*y[t]==Piecewise[{{0,0<=t<1},{1,1<=t<2},{-1,t>=2}}],{y[0]==3,y'[0]==-1}},y[t],t,IncludeSingularSolutions -> True]