problem Problem 2(j)[k]

Internal problem ID [12318]
Internal ﬁle name [OUTPUT/10971_Monday_October_02_2023_02_47_39_AM_43924496/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(j)[k].
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {4 y^{\prime \prime }-4 y^{\prime }+y=t^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = -12, y^{\prime }\left (0\right ) = 7] \end {align*}

4.11.1 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) &=-1\\ q(t) &={\frac {1}{4}}\\ F &=\frac {t^{2}}{4} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }+\frac {y}{4} = \frac {t^{2}}{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 )-4 s Y \left (s \right )+4 y \left (0\right )+Y \left (s \right ) = \frac {2}{s^{3}}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=-12\\ y'(0) &=7 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )-76+48 s -4 s Y \left (s \right )+Y \left (s \right ) = \frac {2}{s^{3}} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = -\frac {2 \left (24 s^{4}-38 s^{3}-1\right )}{s^{3} \left (4 s^{2}-4 s +1\right )} \end {align*}

Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{s^{3}}+\frac {8}{s^{2}}+\frac {24}{s}+\frac {17}{\left (s -\frac {1}{2}\right )^{2}}-\frac {36}{s -\frac {1}{2}} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{s^{3}}\right ) &= t^{2}\\ \mathcal {L}^{-1}\left (\frac {8}{s^{2}}\right ) &= 8 t\\ \mathcal {L}^{-1}\left (\frac {24}{s}\right ) &= 24\\ \mathcal {L}^{-1}\left (\frac {17}{\left (s -\frac {1}{2}\right )^{2}}\right ) &= 17 t \,{\mathrm e}^{\frac {t}{2}}\\ \mathcal {L}^{-1}\left (-\frac {36}{s -\frac {1}{2}}\right ) &= -36 \,{\mathrm e}^{\frac {t}{2}} \end {align*}

Adding the above results and simplifying gives \[ y=t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\right ) \] Simplifying the solution gives \[ y = t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\right ) \] Veriﬁed OK.

4.11.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 \frac {d}{d t}y^{\prime }-4 y^{\prime }+y=t^{2}, y \left (0\right )=-12, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=7\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=y^{\prime }-\frac {y}{4}+\frac {t^{2}}{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} \\ {} & {} & \frac {d}{d t}y^{\prime }-y^{\prime }+\frac {y}{4}=\frac {t^{2}}{4} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-r +\frac {1}{4}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (2 r -1\right )^{2}}{4}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\frac {1}{2} \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\frac {t}{2}} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{\frac {t}{2}} \\ \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}^{\frac {t}{2}}+c_{2} t \,{\mathrm e}^{\frac {t}{2}}+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 {t^{2}}{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}^{\frac {t}{2}} & t \,{\mathrm e}^{\frac {t}{2}} \\ \frac {{\mathrm e}^{\frac {t}{2}}}{2} & {\mathrm e}^{\frac {t}{2}}+\frac {t \,{\mathrm e}^{\frac {t}{2}}}{2} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )={\mathrm e}^{t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\frac {{\mathrm e}^{\frac {t}{2}} \left (\int t^{3} {\mathrm e}^{-\frac {t}{2}}d t -\left (\int t^{2} {\mathrm e}^{-\frac {t}{2}}d t \right ) t \right )}{4} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=t^{2}+8 t +24 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{\frac {t}{2}}+c_{2} t \,{\mathrm e}^{\frac {t}{2}}+t^{2}+8 t +24 \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{\frac {t}{2}}+c_{2} t {\mathrm e}^{\frac {t}{2}}+t^{2}+8 t +24 \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=-12 \\ {} & {} & -12=c_{1} +24 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1} {\mathrm e}^{\frac {t}{2}}}{2}+c_{2} {\mathrm e}^{\frac {t}{2}}+\frac {c_{2} t \,{\mathrm e}^{\frac {t}{2}}}{2}+2 t +8 \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=7 \\ {} & {} & 7=\frac {c_{1}}{2}+c_{2} +8 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-36, c_{2} =17\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (17 t -36\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`

\[ y \left (t \right ) = t^{2}+8 t +24+{\mathrm e}^{\frac {t}{2}} \left (-36+17 t \right ) \]

4.11 problem Problem 2(j)[k]

4.11.1 Existence and uniqueness analysis

4.11.2 Maple step by step solution