problem Problem 14

Internal problem ID [2852]
Internal ﬁle name [OUTPUT/2344_Sunday_June_05_2022_02_59_58_AM_93310471/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.4. page 689
Problem number: Problem 14.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_laplace", "exact linear second order ode", "second_order_integrable_as_is", "second_order_ode_missing_y", "second_order_linear_constant_coeﬀ"

\[ \boxed {y^{\prime \prime }-2 y^{\prime }=30 \,{\mathrm e}^{-3 t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}

13.14.1 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) &=-2\\ q(t) &=0\\ F &=30 \,{\mathrm e}^{-3 t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-2 y^{\prime } = 30 \,{\mathrm e}^{-3 t} \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 ) = \frac {30}{s +3}\tag {1} \end {align*}

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

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+2-s -2 s Y \left (s \right ) = \frac {30}{s +3} \end {align*}

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

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

Adding the above results and simplifying gives \[ y=3 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{-3 t}-4 \] Simplifying the solution gives \[ y = \left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \] Veriﬁed OK.

13.14.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y^{\prime }-2 y^{\prime }=30 \,{\mathrm e}^{-3 t}, y \left (0\right )=1, 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}-2 r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r -2\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (0, 2\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{2 t} \\ \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} +c_{2} {\mathrm e}^{2 t}+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 )=30 \,{\mathrm e}^{-3 t}\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} 1 & {\mathrm e}^{2 t} \\ 0 & 2 \,{\mathrm e}^{2 t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=2 \,{\mathrm e}^{2 t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-15 \left (\int {\mathrm e}^{-3 t}d t \right )+15 \,{\mathrm e}^{2 t} \left (\int {\mathrm e}^{-5 t}d t \right ) \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=2 \,{\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y=c_{1} +c_{2} {\mathrm e}^{2 t}+2 \,{\mathrm e}^{-3 t} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} +c_{2} {\mathrm e}^{2 t}+2 {\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} +c_{2} +2 \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 c_{2} {\mathrm e}^{2 t}-6 \,{\mathrm e}^{-3 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=2 c_{2} -6 \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-4, c_{2} =3\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = 2*_b(_a)+30*exp(-3*_a), _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful`

\[ y \left (t \right ) = \left (3 \,{\mathrm e}^{5 t}-4 \,{\mathrm e}^{3 t}+2\right ) {\mathrm e}^{-3 t} \]

13.14 problem Problem 14

13.14.1 Existence and uniqueness analysis

13.14.2 Maple step by step solution