Internal problem ID [12285]
Internal file name [OUTPUT/10938_Saturday_September_30_2023_08_26_32_PM_31591721/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 5.5 Laplace transform. Homogeneous equations. Problems page 357
Problem number: Problem 5.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y^{\prime \prime }-4 y^{\prime }+5 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 3] \end {align*}
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 &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-4 y^{\prime }+5 y = 0 \end {align*}
The domain of \(p(t)=-4\) is \[
\{-\infty 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 ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y'(0) &=3 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-3-4 s Y \left (s \right )+5 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {3}{s^{2}-4 s +5} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= -\frac {3 i}{2 \left (s -2-i\right )}+\frac {3 i}{2 \left (s -2+i\right )} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {3 i}{2 \left (s -2-i\right )}\right ) &= -\frac {3 i {\mathrm e}^{\left (2+i\right ) t}}{2}\\ \mathcal {L}^{-1}\left (\frac {3 i}{2 \left (s -2+i\right )}\right ) &= \frac {3 i {\mathrm e}^{\left (2-i\right ) t}}{2} \end {align*}
Adding the above results and simplifying gives \[ y=3 \,{\mathrm e}^{2 t} \sin \left (t \right ) \] Simplifying the solution gives \[
y = 3 \,{\mathrm e}^{2 t} \sin \left (t \right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 3 \,{\mathrm e}^{2 t} \sin \left (t \right ) \\
\end{align*} Verification of solutions
\[
y = 3 \,{\mathrm e}^{2 t} \sin \left (t \right )
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-4 y^{\prime }+5 y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of 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 ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the 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 ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{2 t} \cos \left (t \right )+c_{2} {\mathrm e}^{2 t} \sin \left (t \right ) \\ \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 ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{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 ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=3 \\ {} & {} & 3=2 c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =3\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=3 \,{\mathrm e}^{2 t} \sin \left (t \right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=3 \,{\mathrm e}^{2 t} \sin \left (t \right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.485 (sec). Leaf size: 12
\[
y \left (t \right ) = 3 \,{\mathrm e}^{2 t} \sin \left (t \right )
\]
✓ Solution by Mathematica
Time used: 0.027 (sec). Leaf size: 14
\[
y(t)\to 3 e^{2 t} \sin (t)
\]
3.4.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful`
dsolve([diff(y(t),t$2)-4*diff(y(t),t)+5*y(t)=0,y(0) = 0, D(y)(0) = 3],y(t), singsol=all)
DSolve[{y''[t]-4*y'[t]+5*y[t]==0,{y[0]==0,y'[0]==3}},y[t],t,IncludeSingularSolutions -> True]