problem 36

Internal problem ID [6696]
Internal ﬁle name [OUTPUT/5944_Sunday_June_05_2022_04_03_18_PM_79239426/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. 7.4.1 DERIVATIVES OF A TRANSFORM. Page 309
Problem number: 36.
ODE order: 2.
ODE degree: 1.

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

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

7.9.1 Existence and uniqueness analysis

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

Where here \begin {align*} p(t) &=0\\ q(t) &=1\\ F &=\sin \left (t \right ) \left (1+t \right ) \end {align*}

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

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

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

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

Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{4 \left (s -i\right )^{3}}-\frac {1}{4 \left (s +i\right )^{3}}+\frac {-\frac {1}{4}-\frac {i}{8}}{\left (s -i\right )^{2}}+\frac {-\frac {1}{4}+\frac {i}{8}}{\left (s +i\right )^{2}}-\frac {i}{4 \left (s -i\right )}+\frac {i}{4 s +4 i} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s -i\right )^{3}}\right ) &= -\frac {t^{2} {\mathrm e}^{i t}}{8}\\ \mathcal {L}^{-1}\left (-\frac {1}{4 \left (s +i\right )^{3}}\right ) &= -\frac {t^{2} {\mathrm e}^{-i t}}{8}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{4}-\frac {i}{8}}{\left (s -i\right )^{2}}\right ) &= \left (-\frac {1}{4}-\frac {i}{8}\right ) t \,{\mathrm e}^{i t}\\ \mathcal {L}^{-1}\left (\frac {-\frac {1}{4}+\frac {i}{8}}{\left (s +i\right )^{2}}\right ) &= \left (-\frac {1}{4}+\frac {i}{8}\right ) t \,{\mathrm e}^{-i t}\\ \mathcal {L}^{-1}\left (-\frac {i}{4 \left (s -i\right )}\right ) &= -\frac {i {\mathrm e}^{i t}}{4}\\ \mathcal {L}^{-1}\left (\frac {i}{4 s +4 i}\right ) &= \frac {i {\mathrm e}^{-i t}}{4} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {\left (t +2\right ) \left (\sin \left (t \right )-\cos \left (t \right ) t \right )}{4} \] Simplifying the solution gives \[ y = -\frac {\left (-\sin \left (t \right )+\cos \left (t \right ) t \right ) \left (t +2\right )}{4} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (-\sin \left (t \right )+\cos \left (t \right ) t \right ) \left (t +2\right )}{4} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {\left (-\sin \left (t \right )+\cos \left (t \right ) t \right ) \left (t +2\right )}{4} \] Veriﬁed OK.

7.9.2 Maple step by step solution

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

\[ y \left (t \right ) = -\frac {\left (-\sin \left (t \right )+\cos \left (t \right ) t \right ) \left (t +2\right )}{4} \]

7.9 problem 36

7.9.1 Existence and uniqueness analysis

7.9.2 Maple step by step solution