3.7 problem Problem 8

Internal problem ID [12288]
Internal file name [OUTPUT/10941_Saturday_September_30_2023_08_26_33_PM_50385180/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 8.
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, _missing_x]]

\[ \boxed {y^{\prime \prime }+3 y^{\prime }+2 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 3] \end {align*}

3.7.1 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 &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+3 y^{\prime }+2 y = 0 \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 ) = 0\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y'(0) &=3 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-9-2 s +3 s Y \left (s \right )+2 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s +9}{s^{2}+3 s +2} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {7}{s +1}-\frac {5}{s +2} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {7}{s +1}\right ) &= 7 \,{\mathrm e}^{-t}\\ \mathcal {L}^{-1}\left (-\frac {5}{s +2}\right ) &= -5 \,{\mathrm e}^{-2 t} \end {align*}

Adding the above results and simplifying gives \[ y=7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \] Simplifying the solution gives \[ y = 7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = 7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \] Verified OK.

3.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+3 y^{\prime }+2 y=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3\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}+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 ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the 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 ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{-t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=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} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=-2 c_{1} -c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-5, c_{2} =7\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \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: 4.562 (sec). Leaf size: 17

dsolve([diff(y(t),t$2)+3*diff(y(t),t)+2*y(t)=0,y(0) = 2, D(y)(0) = 3],y(t), singsol=all)
 

\[ y \left (t \right ) = 7 \,{\mathrm e}^{-t}-5 \,{\mathrm e}^{-2 t} \]

✓ Solution by Mathematica

Time used: 0.023 (sec). Leaf size: 18

DSolve[{y''[t]+3*y'[t]+2*y[t]==0,{y[0]==2,y'[0]==3}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to e^{-2 t} \left (7 e^t-5\right ) \]