14.12 problem Problem 38

Internal problem ID [2878]
Internal file name [OUTPUT/2370_Sunday_June_05_2022_03_01_48_AM_51361726/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.7. page 704
Problem number: Problem 38.
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]]

\[ \boxed {y^{\prime \prime }+3 y^{\prime }+2 y=-10 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}

14.12.1 Existence and uniqueness analysis

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 &=-10 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }+2 y = -10 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\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 {10 \,{\mathrm e}^{-\frac {s \pi }{4}}}{s^{2}+1}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=1\\ 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 +3 s Y \left (s \right )+2 Y \left (s \right ) = \frac {10 \,{\mathrm e}^{-\frac {s \pi }{4}}}{s^{2}+1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{3}+3 s^{2}+10 \,{\mathrm e}^{-\frac {s \pi }{4}}+s +3}{\left (s^{2}+1\right ) \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 {s^{3}+3 s^{2}+10 \,{\mathrm e}^{-\frac {s \pi }{4}}+s +3}{\left (s^{2}+1\right ) \left (s^{2}+3 s +2\right )}\right )\\ &= -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t}+\frac {\sqrt {2}\, \left (-2 \sqrt {2}\, {\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \sqrt {2}\, {\mathrm e}^{-t +\frac {\pi }{4}}-4 \cos \left (t \right )-2 \sin \left (t \right )\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right )}{2} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} & t <\frac {\pi }{4} \\ -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t}+\frac {\sqrt {2}\, \left (-2 \sqrt {2}\, {\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \sqrt {2}\, {\mathrm e}^{-t +\frac {\pi }{4}}-4 \cos \left (t \right )-2 \sin \left (t \right )\right )}{2} & \frac {\pi }{4}\le t \end {array}\right . \] Simplifying the solution gives \[ y = -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t}-\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{4} \\ 2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}-5 \,{\mathrm e}^{-t +\frac {\pi }{4}}+\sqrt {2}\, \left (\sin \left (t \right )+2 \cos \left (t \right )\right ) & \frac {\pi }{4}\le t \end {array}\right .\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t}-\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{4} \\ 2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}-5 \,{\mathrm e}^{-t +\frac {\pi }{4}}+\sqrt {2}\, \left (\sin \left (t \right )+2 \cos \left (t \right )\right ) & \frac {\pi }{4}\le t \end {array}\right .\right ) \\ \end{align*}

Verification of solutions

\[ y = -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t}-\left (\left \{\begin {array}{cc} 0 & t <\frac {\pi }{4} \\ 2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}-5 \,{\mathrm e}^{-t +\frac {\pi }{4}}+\sqrt {2}\, \left (\sin \left (t \right )+2 \cos \left (t \right )\right ) & \frac {\pi }{4}\le t \end {array}\right .\right ) \] Verified OK.

14.12.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+3 y^{\prime }+2 y=-10 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\right ), y \left (0\right )=1, 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}+3 r +2=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +2\right ) \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, -1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-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}^{-2 t}+c_{2} {\mathrm e}^{-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 )=-10 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\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}^{-2 t} & {\mathrm e}^{-t} \\ -2 \,{\mathrm e}^{-2 t} & -{\mathrm e}^{-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 )=10 \,{\mathrm e}^{-2 t} \left (\int \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\right ) {\mathrm e}^{2 t}d t \right )-10 \,{\mathrm e}^{-t} \left (\int \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \cos \left (t +\frac {\pi }{4}\right ) {\mathrm e}^{t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \left (-2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \,{\mathrm e}^{-t +\frac {\pi }{4}}-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}+\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \left (-2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \,{\mathrm e}^{-t +\frac {\pi }{4}}-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t}+\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \left (-2 {\mathrm e}^{-2 t +\frac {\pi }{2}}+5 {\mathrm e}^{-t +\frac {\pi }{4}}-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}-c_{2} {\mathrm e}^{-t}+\mathit {Dirac}\left (t -\frac {\pi }{4}\right ) \left (-2 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \,{\mathrm e}^{-t +\frac {\pi }{4}}-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right )+\mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) \left (4 \,{\mathrm e}^{-2 t +\frac {\pi }{2}}-5 \,{\mathrm e}^{-t +\frac {\pi }{4}}+\sin \left (t +\frac {\pi }{4}\right )-3 \cos \left (t +\frac {\pi }{4}\right )\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-2 c_{1} -c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-1, c_{2} =2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )-2 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-t +\frac {\pi }{4}}-{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (-\cos \left (t +\frac {\pi }{4}\right )-3 \sin \left (t +\frac {\pi }{4}\right )\right ) \mathit {Heaviside}\left (t -\frac {\pi }{4}\right )-2 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-2 t +\frac {\pi }{2}}+5 \mathit {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-t +\frac {\pi }{4}}-{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} \end {array} \]

Maple trace

`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`
 

✓ Solution by Maple

Time used: 2.375 (sec). Leaf size: 63

dsolve([diff(y(t),t$2)+3*diff(y(t),t)+2*y(t)=10*Heaviside(t-Pi/4)*sin(t-Pi/4),y(0) = 1, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = -2 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{\frac {\pi }{2}-2 t}+5 \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right ) {\mathrm e}^{-t +\frac {\pi }{4}}-2 \sqrt {2}\, \left (\cos \left (t \right )+\frac {\sin \left (t \right )}{2}\right ) \operatorname {Heaviside}\left (t -\frac {\pi }{4}\right )-{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} \]

✓ Solution by Mathematica

Time used: 0.143 (sec). Leaf size: 87

DSolve[{y''[t]+3*y'[t]+2*y[t]==10*UnitStep[t-Pi/4]*Sin[t-Pi/4],{y[0]==1,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-2 t} \left (-1+2 e^t\right ) & 4 t\leq \pi \\ -e^{-2 t} \left (2 \sqrt {2} e^{2 t} \cos (t)-2 e^t-5 e^{t+\frac {\pi }{4}}+\sqrt {2} e^{2 t} \sin (t)+2 e^{\pi /2}+1\right ) & \text {True} \\ \end {array} \\ \end {array} \]