Internal problem ID [6678]
Internal file name [OUTPUT/5926_Sunday_June_05_2022_04_01_52_PM_95041200/index.tex]
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: 31.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }+2 y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 2, y^{\prime }\left (0\right ) = 2] \end {align*}
Since initial condition \(y(0)\) is not at zero, then let \begin {align*} y(0) &= c_{1} \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 {align*} s^{2} Y \left (s \right )-y^{\prime }\left (0\right )-s y \left (0\right )+2 s Y \left (s \right )-2 y \left (0\right )+Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=2 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-2-s c_{1} +2 s Y \left (s \right )-2 c_{1} +Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} +2 c_{1} +2}{s^{2}+2 s +1} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {c_{1}}{s +1}+\frac {c_{1} +2}{\left (s +1\right )^{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {c_{1}}{s +1}\right ) &= {\mathrm e}^{-t} c_{1}\\ \mathcal {L}^{-1}\left (\frac {c_{1} +2}{\left (s +1\right )^{2}}\right ) &= \left (c_{1} +2\right ) t \,{\mathrm e}^{-t} \end {align*}
Adding the above results and simplifying gives \[ y={\mathrm e}^{-t} \left (2 t +c_{1} \left (t +1\right )\right ) \] Since one initial condition \(y(1)\) is not at zero then we need to setup one equation to solve for \(c_{1}\). Using the above solution then \begin {align*} 2 &= {\mathrm e}^{-1} \left (2 c_{1} +2\right ) \end {align*}
Solving the above for \(c_{1}\) gives Solving gives \begin {align*} c_{1} &= -\left ({\mathrm e}^{-1}-1\right ) {\mathrm e} \end {align*}
Subtituting this in the solution obtained above gives \begin {align*} y &= {\mathrm e}^{-t} \left (2 t -\left ({\mathrm e}^{-1}-1\right ) {\mathrm e} \left (t +1\right )\right )\\ &= {\mathrm e}^{-t} \left (t -1+{\mathrm e} t +{\mathrm e}\right ) \end {align*}
Simplifying the solution gives \[ y = {\mathrm e}^{-t} \left (t -1+{\mathrm e} t +{\mathrm e}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-t} \left (t -1+{\mathrm e} t +{\mathrm e}\right ) \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-t} \left (t -1+{\mathrm e} t +{\mathrm e}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }+y=0, y \left (1\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+2 r +1=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r +1\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =-1 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-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}^{-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 ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t} c_{1} +c_{2} t \,{\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y={\mathrm e}^{-t} c_{1} +c_{2} t {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=2 \\ {} & {} & 2={\mathrm e}^{-1} c_{1} +c_{2} {\mathrm e}^{-1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{-t}-c_{2} t \,{\mathrm e}^{-t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {{\mathrm e}^{-1}-1}{{\mathrm e}^{-1}}, c_{2} =\frac {1+{\mathrm e}^{-1}}{{\mathrm e}^{-1}}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t +1} \left (t +1\right )+{\mathrm e}^{-t} \left (t -1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t +1} \left (t +1\right )+{\mathrm e}^{-t} \left (t -1\right ) \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 1.547 (sec). Leaf size: 26
dsolve([diff(y(t),t$2)+2*diff(y(t),t)+y(t)=0,y(1) = 2, D(y)(0) = 2],y(t), singsol=all)
\[ y \left (t \right ) = {\mathrm e}^{-t} \left (t -1+{\mathrm e} t +{\mathrm e}\right ) \]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 18
DSolve[{y''[t]+2*y'[t]+y[t]==0,{y[1]==2,y'[0]==2}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-t} (e t+t+e-1) \]