problem Problem 24.32

Internal problem ID [5209]
Internal ﬁle name [OUTPUT/4702_Sunday_June_05_2022_03_03_39_PM_85824528/index.tex]

Book: Schaums Outline Diﬀerential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems. page 248
Problem number: Problem 24.32.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+2 y^{\prime }+5 y=3 \,{\mathrm e}^{-2 x}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 1] \end {align*}

5.10.1 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) &=5\\ F &=3 \,{\mathrm e}^{-2 x} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+2 y^{\prime }+5 y = 3 \,{\mathrm e}^{-2 x} \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 )+5 Y \left (s \right ) = \frac {3}{s +2}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-3-s +2 s Y \left (s \right )+5 Y \left (s \right ) = \frac {3}{s +2} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}+5 s +9}{\left (s +2\right ) \left (s^{2}+2 s +5\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {3}{5 \left (s +2\right )}+\frac {\frac {1}{5}-\frac {13 i}{20}}{s +1-2 i}+\frac {\frac {1}{5}+\frac {13 i}{20}}{s +1+2 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {3}{5 \left (s +2\right )}\right ) &= \frac {3 \,{\mathrm e}^{-2 x}}{5}\\ \mathcal {L}^{-1}\left (\frac {\frac {1}{5}-\frac {13 i}{20}}{s +1-2 i}\right ) &= \left (\frac {1}{5}-\frac {13 i}{20}\right ) {\mathrm e}^{\left (-1+2 i\right ) x}\\ \mathcal {L}^{-1}\left (\frac {\frac {1}{5}+\frac {13 i}{20}}{s +1+2 i}\right ) &= \left (\frac {1}{5}+\frac {13 i}{20}\right ) {\mathrm e}^{\left (-1-2 i\right ) x} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \] Simplifying the solution gives \[ y = \frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \] Veriﬁed OK.

5.10.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }+2 y^{\prime }+5 y=3 \,{\mathrm e}^{-2 x}, y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\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 +5=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-16}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-2 \,\mathrm {I}, -1+2 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-x} \cos \left (2 x \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )={\mathrm e}^{-x} \sin \left (2 x \right ) \\ \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} \cos \left (2 x \right )+c_{2} {\mathrm e}^{-x} \sin \left (2 x \right )+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 )=3 \,{\mathrm e}^{-2 x}\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} \cos \left (2 x \right ) & {\mathrm e}^{-x} \sin \left (2 x \right ) \\ -{\mathrm e}^{-x} \cos \left (2 x \right )-2 \,{\mathrm e}^{-x} \sin \left (2 x \right ) & -{\mathrm e}^{-x} \sin \left (2 x \right )+2 \,{\mathrm e}^{-x} \cos \left (2 x \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=2 \,{\mathrm e}^{-2 x} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\frac {3 \,{\mathrm e}^{-x} \left (\cos \left (2 x \right ) \left (\int {\mathrm e}^{-x} \sin \left (2 x \right )d x \right )-\sin \left (2 x \right ) \left (\int {\mathrm e}^{-x} \cos \left (2 x \right )d x \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=\frac {3 \,{\mathrm e}^{-2 x}}{5} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-x} \cos \left (2 x \right )+c_{2} {\mathrm e}^{-x} \sin \left (2 x \right )+\frac {3 \,{\mathrm e}^{-2 x}}{5} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-x} \cos \left (2 x \right )+c_{2} {\mathrm e}^{-x} \sin \left (2 x \right )+\frac {3 {\mathrm e}^{-2 x}}{5} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +\frac {3}{5} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-c_{1} {\mathrm e}^{-x} \cos \left (2 x \right )-2 c_{1} {\mathrm e}^{-x} \sin \left (2 x \right )-c_{2} {\mathrm e}^{-x} \sin \left (2 x \right )+2 c_{2} {\mathrm e}^{-x} \cos \left (2 x \right )-\frac {6 \,{\mathrm e}^{-2 x}}{5} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-c_{1} -\frac {6}{5}+2 c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {2}{5}, c_{2} =\frac {13}{10}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \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`

\[ y \left (x \right ) = \frac {3 \,{\mathrm e}^{-2 x}}{5}+\frac {{\mathrm e}^{-x} \left (4 \cos \left (2 x \right )+13 \sin \left (2 x \right )\right )}{10} \]

\[ y(x)\to \frac {1}{10} e^{-2 x} \left (13 e^x \sin (2 x)+4 e^x \cos (2 x)+6\right ) \]

5.10 problem Problem 24.32

5.10.1 Existence and uniqueness analysis

5.10.2 Maple step by step solution