3.4 problem 11

Internal problem ID [836]
Internal file name [OUTPUT/836_Sunday_June_05_2022_01_50_48_AM_37859137/index.tex]

Book: Elementary differential equations and boundary value problems, 11th ed., Boyce, DiPrima, Meade
Section: Chapter 6.2, The Laplace Transform. Solution of Initial Value Problems. page 255
Problem number: 11.
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]]

\[ \boxed {y^{\prime \prime }-2 y^{\prime }+4 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 0] \end {align*}

3.4.1 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) &=4\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime }+4 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 )+4 Y \left (s \right ) = 0\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+4-2 s -2 s Y \left (s \right )+4 Y \left (s \right ) = 0 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {2 s -4}{s^{2}-2 s +4} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {1+\frac {i \sqrt {3}}{3}}{s -1-i \sqrt {3}}+\frac {1-\frac {i \sqrt {3}}{3}}{s -1+i \sqrt {3}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1+\frac {i \sqrt {3}}{3}}{s -1-i \sqrt {3}}\right ) &= \frac {\left (i \sqrt {3}+3\right ) {\mathrm e}^{\left (1+i \sqrt {3}\right ) t}}{3}\\ \mathcal {L}^{-1}\left (\frac {1-\frac {i \sqrt {3}}{3}}{s -1+i \sqrt {3}}\right ) &= \frac {\left (-i \sqrt {3}+3\right ) {\mathrm e}^{\left (1-i \sqrt {3}\right ) t}}{3} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {2 \,{\mathrm e}^{t} \left (3 \cos \left (\sqrt {3}\, t \right )-\sin \left (\sqrt {3}\, t \right ) \sqrt {3}\right )}{3} \] Simplifying the solution gives \[ y = -\frac {2 \,{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}-3 \cos \left (\sqrt {3}\, t \right )\right )}{3} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \,{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}-3 \cos \left (\sqrt {3}\, t \right )\right )}{3} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = -\frac {2 \,{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}-3 \cos \left (\sqrt {3}\, t \right )\right )}{3} \] Verified OK.

3.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }+4 y=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\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 +4=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-12}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-\mathrm {I} \sqrt {3}, 1+\mathrm {I} \sqrt {3}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=\sin \left (\sqrt {3}\, 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}^{t} \cos \left (\sqrt {3}\, t \right )+c_{2} \sin \left (\sqrt {3}\, t \right ) {\mathrm e}^{t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )+c_{2} \sin \left (\sqrt {3}\, t \right ) {\mathrm e}^{t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{t} \cos \left (\sqrt {3}\, t \right )-c_{1} {\mathrm e}^{t} \sin \left (\sqrt {3}\, t \right ) \sqrt {3}+c_{2} \sqrt {3}\, \cos \left (\sqrt {3}\, t \right ) {\mathrm e}^{t}+c_{2} \sin \left (\sqrt {3}\, t \right ) {\mathrm e}^{t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +\sqrt {3}\, c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =-\frac {2 \sqrt {3}}{3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\frac {2 \,{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}-3 \cos \left (\sqrt {3}\, t \right )\right )}{3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {2 \,{\mathrm e}^{t} \left (\sin \left (\sqrt {3}\, t \right ) \sqrt {3}-3 \cos \left (\sqrt {3}\, t \right )\right )}{3} \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: 0.547 (sec). Leaf size: 28

dsolve([diff(y(t),t$2)-2*diff(y(t),t)+4*y(t)=0,y(0) = 2, D(y)(0) = 0],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {2 \left (\sqrt {3}\, \sin \left (\sqrt {3}\, t \right )-3 \cos \left (\sqrt {3}\, t \right )\right ) {\mathrm e}^{t}}{3} \]

✓ Solution by Mathematica

Time used: 0.02 (sec). Leaf size: 37

DSolve[{y''[t]-2*y'[t]+4*y[t]==0,{y[0]==2,y'[0]==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\frac {2}{3} e^t \left (\sqrt {3} \sin \left (\sqrt {3} t\right )-3 \cos \left (\sqrt {3} t\right )\right ) \]