problem 13

Internal problem ID [12804]
Internal ﬁle name [OUTPUT/11457_Saturday_November_04_2023_08_47_24_AM_52105283/index.tex]

Book: Ordinary Diﬀerential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 5. The Laplace Transform Method. Exercises 5.3, page 255
Problem number: 13.
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"

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

14.7.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(x) &=-2\\ q(x) &=1\\ F &=2 \sin \left (x \right ) \end {align*}

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {2 \left (s^{3}-2 s^{2}+s -3\right )}{\left (s^{2}+1\right ) \left (s^{2}-2 s +1\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= -\frac {3}{s -1}+\frac {3}{\left (s -1\right )^{2}}+\frac {1}{2 s -2 i}+\frac {1}{2 s +2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {3}{s -1}\right ) &= -3 \,{\mathrm e}^{x}\\ \mathcal {L}^{-1}\left (\frac {3}{\left (s -1\right )^{2}}\right ) &= 3 x \,{\mathrm e}^{x}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s -2 i}\right ) &= \frac {{\mathrm e}^{i x}}{2}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s +2 i}\right ) &= \frac {{\mathrm e}^{-i x}}{2} \end {align*}

Adding the above results and simplifying gives \[ y=\cos \left (x \right )+3 \left (x -1\right ) {\mathrm e}^{x} \] Simplifying the solution gives \[ y = \left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \] Veriﬁed OK.

14.7.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{\prime }+y=2 \sin \left (x \right ), y \left (0\right )=-2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \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 homogeneous 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 homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{x} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \,{\mathrm e}^{x} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \left (\int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right )+y_{2}\left (x \right ) \left (\int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x \right ), f \left (x \right )=2 \sin \left (x \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} {\mathrm e}^{x} & x \,{\mathrm e}^{x} \\ {\mathrm e}^{x} & {\mathrm e}^{x}+x \,{\mathrm e}^{x} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )={\mathrm e}^{2 x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=2 \,{\mathrm e}^{x} \left (-\left (\int x \,{\mathrm e}^{-x} \sin \left (x \right )d x \right )+x \left (\int {\mathrm e}^{-x} \sin \left (x \right )d x \right )\right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\cos \left (x \right ) \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}+\cos \left (x \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{x}+c_{2} x {\mathrm e}^{x}+\cos \left (x \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-2 \\ {} & {} & -2=1+c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{1} {\mathrm e}^{x}+c_{2} {\mathrm e}^{x}+c_{2} x \,{\mathrm e}^{x}-\sin \left (x \right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-3, c_{2} =3\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful`

\[ y \left (x \right ) = \left (3 x -3\right ) {\mathrm e}^{x}+\cos \left (x \right ) \]

14.7 problem 13

14.7.1 Existence and uniqueness analysis

14.7.2 Maple step by step solution