problem 29

Internal problem ID [13224]
Internal ﬁle name [OUTPUT/11880_Tuesday_December_05_2023_12_12_41_PM_3851012/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 6. Laplace transform. Section 6.3 page 600
Problem number: 29.
ODE order: 2.
ODE degree: 1.

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

\[ \boxed {y^{\prime \prime }-4 y^{\prime }+5 y=2 \,{\mathrm e}^{t}} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 1] \end {align*}

19.3.1 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) &=5\\ F &=2 \,{\mathrm e}^{t} \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-4 y^{\prime }+5 y = 2 \,{\mathrm e}^{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 )-4 s Y \left (s \right )+4 y \left (0\right )+5 Y \left (s \right ) = \frac {2}{s -1}\tag {1} \end {align*}

But the initial conditions are \begin {align*} y \left (0\right )&=3\\ y'(0) &=1 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )+11-3 s -4 s Y \left (s \right )+5 Y \left (s \right ) = \frac {2}{s -1} \end {align*}

Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3 s^{2}-14 s +13}{\left (s -1\right ) \left (s^{2}-4 s +5\right )} \end {align*}

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

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

Summary

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

Veriﬁcation of solutions

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

19.3.2 Maple step by step solution

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

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

19.3 problem 29

19.3.1 Existence and uniqueness analysis

19.3.2 Maple step by step solution