3.3 problem Problem 4

Internal problem ID [12284]
Internal file name [OUTPUT/10937_Saturday_September_30_2023_08_26_31_PM_64152595/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.5 Laplace transform. Homogeneous equations. Problems page 357
Problem number: Problem 4.
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 (0\right ) = -1, y^{\prime }\left (0\right ) = 2] \end {align*}

3.3.1 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) &=2\\ q(t) &=1\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+y = 0 \end {align*}

The domain of \(p(t)=2\) 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 )+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 )&=-1\\ y'(0) &=2 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+s +2 s Y \left (s \right )+Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {s}{s^{2}+2 s +1} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{\left (s +1\right )^{2}}-\frac {1}{s +1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{\left (s +1\right )^{2}}\right ) &= t \,{\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (-\frac {1}{s +1}\right ) &= -{\mathrm e}^{-t} \end {align*}

Adding the above results and simplifying gives \[ y={\mathrm e}^{-t} \left (t -1\right ) \] Simplifying the solution gives \[ y = {\mathrm e}^{-t} \left (t -1\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-t} \left (t -1\right ) \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = {\mathrm e}^{-t} \left (t -1\right ) \] Verified OK.

3.3.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }+y=0, y \left (0\right )=-1, 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=c_{1} {\mathrm e}^{-t}+c_{2} t \,{\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-t}+c_{2} t {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-t}+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} =-1, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{-t} \left (t -1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\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: 4.469 (sec). Leaf size: 12

dsolve([diff(y(t),t$2)+2*diff(y(t),t)+y(t)=0,y(0) = -1, D(y)(0) = 2],y(t), singsol=all)
 

\[ y \left (t \right ) = {\mathrm e}^{-t} \left (t -1\right ) \]

✓ Solution by Mathematica

Time used: 0.025 (sec). Leaf size: 14

DSolve[{y''[t]+2*y'[t]+y[t]==0,{y[0]==-1,y'[0]==2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-t} (t-1) \]