6.4 problem 24

Internal problem ID [7321]
Book : DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section : CHAPTER 7 THE LAPLACE TRANSFORM. 7.3.1 TRANSLATION ON THE s-AXIS. Page 297
Problem number : 24
Date solved : Thursday, October 17, 2024 at 10:55:54 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} y^{\prime \prime }-4 y^{\prime }+4 y&=t^{3} {\mathrm e}^{2 t} \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0 \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{equation} \tag{1} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )-4 s Y \left (s \right )+4 y \left (0\right )+4 Y \left (s \right ) = \frac {6}{\left (s -2\right )^{4}} \end{equation}

Substituting the initial conditions in above in Eq (1) gives

\[ s^{2} Y \left (s \right )-4 s Y \left (s \right )+4 Y \left (s \right ) = \frac {6}{\left (s -2\right )^{4}} \]

Solving the above equation for \(Y(s)\) results in

\[ Y(s) = \frac {6}{\left (s -2\right )^{4} \left (s^{2}-4 s +4\right )} \]

Taking inverse Laplace transform gives

\[ \mathcal {L}^{-1}\left (\frac {6}{\left (s -2\right )^{4} \left (s^{2}-4 s +4\right )}\right ) = \frac {t^{5} {\mathrm e}^{2 t}}{20} \]

Solving for \(c_1\) and \(c_2\) from the given initial conditions results in

\[ y = \frac {t^{5} {\mathrm e}^{2 t}}{20} \]

Simplifying the solution gives

\[ y = \frac {t^{5} {\mathrm e}^{2 t}}{20} \]

(a) Solution plot
\(y = \frac {t^{5} {\mathrm e}^{2 t}}{20}\)

(b) Slope field plot
\(y^{\prime \prime }-4 y^{\prime }+4 y = t^{3} {\mathrm e}^{2 t}\)

6.4.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d^{2}}{d t^{2}}y \left (t \right )-4 \frac {d}{d t}y \left (t \right )+4 y \left (t \right )=t^{3} {\mathrm e}^{2 t}, y \left (0\right )=0, \left (\frac {d}{d t}y \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =2 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{2 t}+\mathit {C2} t \,{\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 )=t^{3} {\mathrm e}^{2 t}\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} & t \,{\mathrm e}^{2 t} \\ 2 \,{\mathrm e}^{2 t} & {\mathrm e}^{2 t}+2 t \,{\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}^{4 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{2 t} \left (-\left (\int t^{4}d t \right )+\left (\int t^{3}d t \right ) t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{2 t} t^{5}}{20} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{2 t}+\mathit {C2} t \,{\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} t^{5}}{20} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y \left (t \right )=\textit {\_C1} {\mathrm e}^{2 t}+\textit {\_C2} t {\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} t^{5}}{20} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\textit {\_C1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=2 \textit {\_C1} \,{\mathrm e}^{2 t}+\textit {\_C2} \,{\mathrm e}^{2 t}+2 \textit {\_C2} t \,{\mathrm e}^{2 t}+\frac {{\mathrm e}^{2 t} t^{5}}{10}+\frac {{\mathrm e}^{2 t} t^{4}}{4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}y \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=2 \textit {\_C1} +\textit {\_C2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \hspace {3pt}\textrm {and}\hspace {3pt} \textit {\_C2} \\ {} & {} & \left \{\textit {\_C1} =0, \textit {\_C2} =0\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {{\mathrm e}^{2 t} t^{5}}{20} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {{\mathrm e}^{2 t} t^{5}}{20} \end {array} \]

6.4.2 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)] 
-> 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`
 

6.4.3 Maple dsolve solution

Solving time : 0.125 (sec)
Leaf size : 13

dsolve([diff(diff(y(t),t),t)-4*diff(y(t),t)+4*y(t) = t^3*exp(2*t), 
        op([y(0) = 0, D(y)(0) = 0])], 
        y(t),method=laplace)
 

\[ y = \frac {t^{5} {\mathrm e}^{2 t}}{20} \]

6.4.4 Mathematica DSolve solution

Solving time : 0.023 (sec)
Leaf size : 17

DSolve[{D[y[t],{t,2}]-4*D[y[t],t]+4*y[t]==t^3*Exp[2*t],{y[0]==0,Derivative[1][y][0] ==0}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \frac {1}{20} e^{2 t} t^5 \]