problem 15(b)

Internal problem ID [6710]
Internal ﬁle name [OUTPUT/5958_Sunday_June_05_2022_04_04_25_PM_69753260/index.tex]

Book: DIFFERENTIAL EQUATIONS with Boundary Value Problems. DENNIS G. ZILL, WARREN S. WRIGHT, MICHAEL R. CULLEN. Brooks/Cole. Boston, MA. 2013. 8th edition.
Section: CHAPTER 7 THE LAPLACE TRANSFORM. EXERCISES 7.5. Page 315
Problem number: 15(b).
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }+2 y^{\prime }+10 y=\delta \left (t \right )} \] Since no initial conditions are explicitly given, then let \begin {align*} y \left (0\right )&= c_{1} \\ y'(0) &= c_{2} \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 )+10 Y \left (s \right ) = 1\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} +2 s Y \left (s \right )-2 c_{1} +10 Y \left (s \right ) = 1 \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} +2 c_{1} +c_{2} +1}{s^{2}+2 s +10} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {\left (-1+3 i\right ) \left (-\frac {c_{1}}{18}-\frac {c_{2}}{18}-\frac {1}{18}\right )+\frac {4 c_{1}}{9}-\frac {c_{2}}{18}-\frac {1}{18}}{s +1-3 i}+\frac {\left (-1-3 i\right ) \left (-\frac {c_{1}}{18}-\frac {c_{2}}{18}-\frac {1}{18}\right )+\frac {4 c_{1}}{9}-\frac {c_{2}}{18}-\frac {1}{18}}{s +1+3 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {\left (-1+3 i\right ) \left (-\frac {c_{1}}{18}-\frac {c_{2}}{18}-\frac {1}{18}\right )+\frac {4 c_{1}}{9}-\frac {c_{2}}{18}-\frac {1}{18}}{s +1-3 i}\right ) &= \frac {{\mathrm e}^{\left (-1+3 i\right ) t} \left (-i-i c_{2} +\left (3-i\right ) c_{1} \right )}{6}\\ \mathcal {L}^{-1}\left (\frac {\left (-1-3 i\right ) \left (-\frac {c_{1}}{18}-\frac {c_{2}}{18}-\frac {1}{18}\right )+\frac {4 c_{1}}{9}-\frac {c_{2}}{18}-\frac {1}{18}}{s +1+3 i}\right ) &= \frac {{\mathrm e}^{\left (-1-3 i\right ) t} \left (i+i c_{2} +\left (3+i\right ) c_{1} \right )}{6} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{-t} \left (3 c_{1} \cos \left (3 t \right )+\sin \left (3 t \right ) \left (c_{1} +c_{2} +1\right )\right )}{3} \] Simplifying the solution gives \[ y = \frac {\left (c_{1} +c_{2} +1\right ) {\mathrm e}^{-t} \sin \left (3 t \right )}{3}+c_{1} \cos \left (3 t \right ) {\mathrm e}^{-t} \]

Summary

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

Veriﬁcation of solutions

\[ y = \frac {\left (c_{1} +c_{2} +1\right ) {\mathrm e}^{-t} \sin \left (3 t \right )}{3}+c_{1} \cos \left (3 t \right ) {\mathrm e}^{-t} \] Veriﬁed OK.

8.14.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+2 y^{\prime }+10 y=\mathit {Dirac}\left (t \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 +10=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-2\right )\pm \left (\sqrt {-36}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1-3 \,\mathrm {I}, -1+3 \,\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=\cos \left (3 t \right ) {\mathrm e}^{-t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-t} \sin \left (3 t \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (3 t \right ) {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-t} \sin \left (3 t \right )+y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \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 (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \left (\int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right )+y_{2}\left (t \right ) \left (\int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=\mathit {Dirac}\left (t \right )\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} \cos \left (3 t \right ) {\mathrm e}^{-t} & {\mathrm e}^{-t} \sin \left (3 t \right ) \\ -3 \,{\mathrm e}^{-t} \sin \left (3 t \right )-\cos \left (3 t \right ) {\mathrm e}^{-t} & -{\mathrm e}^{-t} \sin \left (3 t \right )+3 \cos \left (3 t \right ) {\mathrm e}^{-t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=3 \,{\mathrm e}^{-2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-t} \sin \left (3 t \right ) \left (\int \mathit {Dirac}\left (t \right )d t \right )}{3} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-t} \sin \left (3 t \right ) \mathit {Heaviside}\left (t \right )}{3} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} \cos \left (3 t \right ) {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-t} \sin \left (3 t \right )+\frac {{\mathrm e}^{-t} \sin \left (3 t \right ) \mathit {Heaviside}\left (t \right )}{3} \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`

\[ y \left (t \right ) = \frac {{\mathrm e}^{-t} \left (3 y \left (0\right ) \cos \left (3 t \right )+\sin \left (3 t \right ) \left (D\left (y \right )\left (0\right )+y \left (0\right )+1\right )\right )}{3} \]

\[ y(t)\to \frac {1}{3} e^{-t} (\theta (t) \sin (3 t)+3 c_2 \cos (3 t)+3 c_1 \sin (3 t)) \]

8.14 problem 15(b)

8.14.1 Maple step by step solution