problem 6(d)

Internal problem ID [11510]
Internal ﬁle name [OUTPUT/10493_Thursday_May_18_2023_04_21_03_AM_81125574/index.tex]

Book: A First Course in Diﬀerential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 3, Laplace transform. Section 3.2.1 Initial value problems. Exercises page 156
Problem number: 6(d).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeﬀ"

\[ \boxed {x^{\prime \prime }-2 x^{\prime }+2 x=0} \] With initial conditions \begin {align*} [x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 1] \end {align*}

15.4.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}

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

Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (x\right ) =Y(s) \end {align*}

Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right ) &= s Y(s) - x \left (0\right )\\ \mathcal {L}\left (x^{\prime \prime }\right ) &= s^2 Y(s) - x'(0) - s x \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 )-x^{\prime }\left (0\right )-s x \left (0\right )-2 s Y \left (s \right )+2 x \left (0\right )+2 Y \left (s \right ) = 0\tag {1} \end {align*}

But the initial conditions are \begin {align*} x \left (0\right )&=0\\ x'(0) &=1 \end {align*}

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

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

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

Summary

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

Veriﬁcation of solutions

\[ x = {\mathrm e}^{t} \sin \left (t \right ) \] Veriﬁed OK.

15.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{\prime \prime }-2 x^{\prime }+2 x=0, x \left (0\right )=0, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-2 r +2=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {2\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1-\mathrm {I}, 1+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{t} \sin \left (t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & x=c_{1} {\mathrm e}^{t} \cos \left (t \right )+c_{2} {\mathrm e}^{t} \sin \left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} {\mathrm e}^{t} \cos \left (t \right )+c_{2} {\mathrm e}^{t} \sin \left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=c_{1} {\mathrm e}^{t} \cos \left (t \right )-c_{1} {\mathrm e}^{t} \sin \left (t \right )+c_{2} {\mathrm e}^{t} \sin \left (t \right )+c_{2} {\mathrm e}^{t} \cos \left (t \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{t} \sin \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x={\mathrm e}^{t} \sin \left (t \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`

15.4 problem 6(d)

15.4.1 Existence and uniqueness analysis

15.4.2 Maple step by step solution