problem Problem 24.31

Internal problem ID [5208]
Internal ﬁle name [OUTPUT/4701_Sunday_June_05_2022_03_03_38_PM_30494857/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 24.31.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+y^{\prime }+y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 4, y^{\prime }\left (0\right ) = -3] \end {align*}

5.9.1 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*}

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

But the initial conditions are \begin {align*} y \left (0\right )&=4\\ y'(0) &=-3 \end {align*}

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

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {4 s +1}{s^{2}+s +1} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {2+\frac {i \sqrt {3}}{3}}{s +\frac {1}{2}-\frac {i \sqrt {3}}{2}}+\frac {2-\frac {i \sqrt {3}}{3}}{s +\frac {1}{2}+\frac {i \sqrt {3}}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2+\frac {i \sqrt {3}}{3}}{s +\frac {1}{2}-\frac {i \sqrt {3}}{2}}\right ) &= \frac {\left (i \sqrt {3}+6\right ) {\mathrm e}^{-\frac {\left (1-i \sqrt {3}\right ) x}{2}}}{3}\\ \mathcal {L}^{-1}\left (\frac {2-\frac {i \sqrt {3}}{3}}{s +\frac {1}{2}+\frac {i \sqrt {3}}{2}}\right ) &= \frac {\left (6-i \sqrt {3}\right ) {\mathrm e}^{-\frac {\left (1+i \sqrt {3}\right ) x}{2}}}{3} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {2 \,{\mathrm e}^{-\frac {x}{2}} \left (6 \cos \left (\frac {\sqrt {3}\, x}{2}\right )-\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )\right )}{3} \] Simplifying the solution gives \[ y = -\frac {2 \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-6 \cos \left (\frac {\sqrt {3}\, x}{2}\right )\right ) {\mathrm e}^{-\frac {x}{2}}}{3} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-6 \cos \left (\frac {\sqrt {3}\, x}{2}\right )\right ) {\mathrm e}^{-\frac {x}{2}}}{3} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {2 \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-6 \cos \left (\frac {\sqrt {3}\, x}{2}\right )\right ) {\mathrm e}^{-\frac {x}{2}}}{3} \] Veriﬁed OK.

5.9.2 Maple step by step solution

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

\[ y \left (x \right ) = -\frac {2 \left (\sqrt {3}\, \sin \left (\frac {\sqrt {3}\, x}{2}\right )-6 \cos \left (\frac {\sqrt {3}\, x}{2}\right )\right ) {\mathrm e}^{-\frac {x}{2}}}{3} \]

\[ y(x)\to -\frac {2}{3} e^{-x/2} \left (\sqrt {3} \sin \left (\frac {\sqrt {3} x}{2}\right )-6 \cos \left (\frac {\sqrt {3} x}{2}\right )\right ) \]

5.9 problem Problem 24.31

5.9.1 Existence and uniqueness analysis

5.9.2 Maple step by step solution