problem Problem 2(b)

Internal problem ID [12309]
Internal ﬁle name [OUTPUT/10962_Saturday_September_30_2023_08_26_37_PM_33751077/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number: Problem 2(b).
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {4 y^{\prime \prime }+16 y^{\prime }+17 y=17 t -1} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 2] \end {align*}

4.2.1 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) &=4\\ q(t) &={\frac {17}{4}}\\ F &=\frac {17 t}{4}-\frac {1}{4} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+4 y^{\prime }+\frac {17 y}{4} = \frac {17 t}{4}-\frac {1}{4} \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*} 4 s^{2} Y \left (s \right )-4 y^{\prime }\left (0\right )-4 s y \left (0\right )+16 s Y \left (s \right )-16 y \left (0\right )+17 Y \left (s \right ) = \frac {17}{s^{2}}-\frac {1}{s}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )+8+4 s +16 s Y \left (s \right )+17 Y \left (s \right ) = \frac {17}{s^{2}}-\frac {1}{s} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {4 s^{3}+8 s^{2}+s -17}{s^{2} \left (4 s^{2}+16 s +17\right )} \end {align*}

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

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

Summary

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

Veriﬁcation of solutions

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

4.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 y^{\prime \prime }+16 y^{\prime }+17 y=17 t -1, y \left (0\right )=-1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-4 y^{\prime }-\frac {17 y}{4}+\frac {17 t}{4}-\frac {1}{4} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+4 y^{\prime }+\frac {17 y}{4}=\frac {17 t}{4}-\frac {1}{4} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+4 r +\frac {17}{4}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {\left (-4\right )\pm \left (\sqrt {-1}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2-\frac {\mathrm {I}}{2}, -2+\frac {\mathrm {I}}{2}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\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} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\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 )=\frac {17 t}{4}-\frac {1}{4}\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} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right ) & {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right ) \\ -2 \,{\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )-\frac {{\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )}{2} & -2 \,{\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )+\frac {{\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {{\mathrm e}^{-4 t}}{2} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=\frac {{\mathrm e}^{-2 t} \left (-\cos \left (\frac {t}{2}\right ) \left (\int \left (17 t -1\right ) \sin \left (\frac {t}{2}\right ) {\mathrm e}^{2 t}d t \right )+\sin \left (\frac {t}{2}\right ) \left (\int \left (17 t -1\right ) \cos \left (\frac {t}{2}\right ) {\mathrm e}^{2 t}d t \right )\right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=t -1 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )+t -1 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )+c_{2} {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )+t -1 \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-1 \\ {} & {} & -1=c_{1} -1 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )-\frac {c_{1} {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )}{2}-2 c_{2} {\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )+\frac {c_{2} {\mathrm e}^{-2 t} \cos \left (\frac {t}{2}\right )}{2}+1 \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=-2 c_{1} +1+\frac {c_{2}}{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-1+t +2 \,{\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-1+t +2 \,{\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\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`

\[ y \left (t \right ) = 2 \,{\mathrm e}^{-2 t} \sin \left (\frac {t}{2}\right )+t -1 \]

4.2 problem Problem 2(b)

4.2.1 Existence and uniqueness analysis

4.2.2 Maple step by step solution