2.5.17 problem 17

Internal problem ID [8404]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 17
Date solved : Sunday, November 10, 2024 at 03:45:05 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }+3 y^{\prime }-4 y&=6 \,{\mathrm e}^{2 t -2} \end{align*}

With initial conditions

\begin{align*} y \left (1\right )&=4\\ y^{\prime }\left (1\right )&=5 \end{align*}

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{equation} \tag{1} 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 )-4 Y \left (s \right ) = \frac {6 \,{\mathrm e}^{-2}}{s -2} \end{equation}

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

\[ s^{2} Y \left (s \right )-c_2 -s c_1 +3 s Y \left (s \right )-3 c_1 -4 Y \left (s \right ) = \frac {6 \,{\mathrm e}^{-2}}{s -2} \]

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

\[ Y(s) = \frac {s^{2} c_1 +s c_1 +c_2 s +6 \,{\mathrm e}^{-2}-6 c_1 -2 c_2}{\left (s -2\right ) \left (s^{2}+3 s -4\right )} \]

Applying partial fractions decomposition results in

\[ Y(s)= \frac {\frac {c_1}{5}-\frac {c_2}{5}+\frac {{\mathrm e}^{-2}}{5}}{s +4}+\frac {{\mathrm e}^{-2}}{s -2}+\frac {\frac {4 c_1}{5}+\frac {c_2}{5}-\frac {6 \,{\mathrm e}^{-2}}{5}}{s -1} \]

The inverse Laplace of each term above is now found, which gives

\begin{align*} \mathcal {L}^{-1}\left (\frac {\frac {c_1}{5}-\frac {c_2}{5}+\frac {{\mathrm e}^{-2}}{5}}{s +4}\right ) &= \frac {\left (c_1 -c_2 +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4 t}}{5} \\ \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-2}}{s -2}\right ) &= {\mathrm e}^{2 t -2} \\ \mathcal {L}^{-1}\left (\frac {\frac {4 c_1}{5}+\frac {c_2}{5}-\frac {6 \,{\mathrm e}^{-2}}{5}}{s -1}\right ) &= \frac {{\mathrm e}^{t} \left (4 c_1 +c_2 -6 \,{\mathrm e}^{-2}\right )}{5} \\ \end{align*}

Adding the above results and simplifying gives

\[ y={\mathrm e}^{2 t -2}+\frac {{\mathrm e}^{t} \left (4 c_1 +c_2 -6 \,{\mathrm e}^{-2}\right )}{5}+\frac {\left (c_1 -c_2 +{\mathrm e}^{-2}\right ) {\mathrm e}^{-4 t}}{5} \]

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

\[ y = {\mathrm e}^{2 t -2}+\frac {{\mathrm e}^{t} \left (4 \left ({\mathrm e}^{-1}+3\right ) {\mathrm e}^{-1}-4 \,{\mathrm e}^{-2}+3 \,{\mathrm e}^{-1}\right )}{5}+\frac {\left (\left ({\mathrm e}^{-1}+3\right ) {\mathrm e}^{-1}-{\mathrm e}^{-2}-3 \,{\mathrm e}^{-1}\right ) {\mathrm e}^{-4 t}}{5} \]

Simplifying the solution gives

\[ y = {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1} \]

(a) Solution plot
\(y = {\mathrm e}^{2 t -2}+3 \,{\mathrm e}^{t -1}\)

(b) Slope field plot
\(y^{\prime \prime }+3 y^{\prime }-4 y = 6 \,{\mathrm e}^{2 t -2}\)

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 )+3 \frac {d}{d t}y \left (t \right )-4 y \left (t \right )=6 \,{\mathrm e}^{2 t -2}, y \left (1\right )=4, \left (\frac {d}{d t}y \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=5\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}+3 r -4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +4\right ) \left (r -1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-4, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-4 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 \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}^{-4 t}+\mathit {C2} \,{\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 )=6 \,{\mathrm e}^{2 t -2}\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}^{-4 t} & {\mathrm e}^{t} \\ -4 \,{\mathrm e}^{-4 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 )=5 \,{\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {6 \left (-{\mathrm e}^{5 t} \left (\int {\mathrm e}^{t -2}d t \right )+\int {\mathrm e}^{6 t -2}d t \right ) {\mathrm e}^{-4 t}}{5} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )={\mathrm e}^{2 t -2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{-4 t}+\mathit {C2} \,{\mathrm e}^{t}+{\mathrm e}^{2 t -2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y \left (t \right )=\textit {\_C1} {\mathrm e}^{-4 t}+\textit {\_C2} {\mathrm e}^{t}+{\mathrm e}^{2 t -2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=4 \\ {} & {} & 4=\textit {\_C1} \,{\mathrm e}^{-4}+\textit {\_C2} \,{\mathrm e}+1 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=-4 \textit {\_C1} \,{\mathrm e}^{-4 t}+\textit {\_C2} \,{\mathrm e}^{t}+2 \,{\mathrm e}^{2 t -2} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d t}y \left (t \right )\right )\bigg | {\mstack {}{_{\left \{t \hiderel {=}1\right \}}}}=5 \\ {} & {} & 5=-4 \textit {\_C1} \,{\mathrm e}^{-4}+\textit {\_C2} \,{\mathrm e}+2 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \hspace {3pt}\textrm {and}\hspace {3pt} \textit {\_C2} \\ {} & {} & \left \{\textit {\_C1} =0, \textit {\_C2} =\frac {3}{{\mathrm e}}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=3 \,{\mathrm e}^{t -1}+{\mathrm e}^{2 t -2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=3 \,{\mathrm e}^{t -1}+{\mathrm e}^{2 t -2} \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)] 
-> 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`
 

Maple dsolve solution

Solving time : 2.088 (sec)
Leaf size : 17

dsolve([diff(diff(y(t),t),t)+3*diff(y(t),t)-4*y(t) = 6*exp(2*t-2), 
        op([y(1) = 4, D(y)(1) = 5])], 
        y(t),method=laplace)
 

\[ y = 3 \,{\mathrm e}^{t -1}+{\mathrm e}^{2 t -2} \]

Mathematica DSolve solution

Solving time : 0.075 (sec)
Leaf size : 18

DSolve[{D[y[t],{t,2}]+3*D[y[t],t]-4*y[t]==6*Exp[2*t-2],{y[1]==4,Derivative[1][y][1]==5}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to e^{t-2} \left (e^t+3 e\right ) \]