problem Problem 2(l)[n]

Internal problem ID [12321]
Internal ﬁle name [OUTPUT/10974_Monday_October_02_2023_02_47_40_AM_13174011/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(l)[n].
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {3 y^{\prime \prime }+5 y^{\prime }-2 y=7 \,{\mathrm e}^{-2 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 0] \end {align*}

4.14.1 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) &={\frac {5}{3}}\\ q(t) &=-{\frac {2}{3}}\\ F &=\frac {7 \,{\mathrm e}^{-2 t}}{3} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }+\frac {5 y^{\prime }}{3}-\frac {2 y}{3} = \frac {7 \,{\mathrm e}^{-2 t}}{3} \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*} 3 s^{2} Y \left (s \right )-3 y^{\prime }\left (0\right )-3 s y \left (0\right )+5 s Y \left (s \right )-5 y \left (0\right )-2 Y \left (s \right ) = \frac {7}{s +2}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} 3 s^{2} Y \left (s \right )-15-9 s +5 s Y \left (s \right )-2 Y \left (s \right ) = \frac {7}{s +2} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {9 s^{2}+33 s +37}{\left (s +2\right ) \left (3 s^{2}+5 s -2\right )} \end {align*}

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

Adding the above results and simplifying gives \[ y=-t \,{\mathrm e}^{-2 t}+3 \,{\mathrm e}^{\frac {t}{3}} \] Simplifying the solution gives \[ y = -\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \] Veriﬁed OK.

4.14.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [3 \frac {d}{d t}y^{\prime }+5 y^{\prime }-2 y=7 \,{\mathrm e}^{-2 t}, y \left (0\right )=3, 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y^{\prime }=-\frac {5 y^{\prime }}{3}+\frac {2 y}{3}+\frac {7 \,{\mathrm e}^{-2 t}}{3} \\ \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 }+\frac {5 y^{\prime }}{3}-\frac {2 y}{3}=\frac {7 \,{\mathrm e}^{-2 t}}{3} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+\frac {5}{3} r -\frac {2}{3}=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \frac {\left (r +2\right ) \left (3 r -1\right )}{3}=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-2, \frac {1}{3}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{3}} \\ \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}+c_{2} {\mathrm e}^{\frac {t}{3}}+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 {7 \,{\mathrm e}^{-2 t}}{3}\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} & {\mathrm e}^{\frac {t}{3}} \\ -2 \,{\mathrm e}^{-2 t} & \frac {{\mathrm e}^{\frac {t}{3}}}{3} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\frac {7 \,{\mathrm e}^{-\frac {5 t}{3}}}{3} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\left (-{\mathrm e}^{\frac {7 t}{3}} \left (\int {\mathrm e}^{-\frac {7 t}{3}}d t \right )+\int 1d t \right ) {\mathrm e}^{-2 t} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=-\frac {\left (3+7 t \right ) {\mathrm e}^{-2 t}}{7} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{\frac {t}{3}}-\frac {\left (3+7 t \right ) {\mathrm e}^{-2 t}}{7} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{-2 t}+c_{2} {\mathrm e}^{\frac {t}{3}}-\frac {\left (3+7 t \right ) {\mathrm e}^{-2 t}}{7} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{1} +c_{2} -\frac {3}{7} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 c_{1} {\mathrm e}^{-2 t}+\frac {c_{2} {\mathrm e}^{\frac {t}{3}}}{3}-{\mathrm e}^{-2 t}+\frac {2 \left (3+7 t \right ) {\mathrm e}^{-2 t}}{7} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-2 c_{1} +\frac {c_{2}}{3}-\frac {1}{7} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {3}{7}, c_{2} =3\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \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`

\[ y \left (t \right ) = -\left (-3 \,{\mathrm e}^{\frac {7 t}{3}}+t \right ) {\mathrm e}^{-2 t} \]

4.14 problem Problem 2(l)[n]

4.14.1 Existence and uniqueness analysis

4.14.2 Maple step by step solution