Internal problem ID [12283]
Internal file name [OUTPUT/10936_Thursday_September_28_2023_01_09_19_AM_72642257/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 3.
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]]
\[ \boxed {4 y^{\prime \prime }-4 y^{\prime }+5 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, 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) &=-1\\ q(t) &={\frac {5}{4}}\\ F &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }-y^{\prime }+\frac {5 y}{4} = 0 \end {align*}
The domain of \(p(t)=-1\) 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*} 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 )+5 Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=2\\ y'(0) &=3 \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 4 s^{2} Y \left (s \right )-4-8 s -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 {8 s +4}{4 s^{2}-4 s +5} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1-i}{s -\frac {1}{2}-i}+\frac {1+i}{s -\frac {1}{2}+i} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1-i}{s -\frac {1}{2}-i}\right ) &= \left (1-i\right ) {\mathrm e}^{\left (\frac {1}{2}+i\right ) t}\\ \mathcal {L}^{-1}\left (\frac {1+i}{s -\frac {1}{2}+i}\right ) &= \left (1+i\right ) {\mathrm e}^{\left (\frac {1}{2}-i\right ) t} \end {align*}
Adding the above results and simplifying gives \[ y=2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right ) \] Simplifying the solution gives \[
y = 2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right ) \\
\end{align*} Verification of solutions
\[
y = 2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right )
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [4 y^{\prime \prime }-4 y^{\prime }+5 y=0, y \left (0\right )=2, 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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=y^{\prime }-\frac {5 y}{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} \\ {} & {} & y^{\prime \prime }-y^{\prime }+\frac {5 y}{4}=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-r +\frac {5}{4}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {1\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\frac {1}{2}-\mathrm {I}, \frac {1}{2}+\mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{\frac {t}{2}} \cos \left (t \right ) \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{\frac {t}{2}} \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}^{\frac {t}{2}} \cos \left (t \right )+c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (t \right ) \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{\frac {t}{2}} \cos \left (t \right )+c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (t \right ) \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {c_{1} {\mathrm e}^{\frac {t}{2}} \cos \left (t \right )}{2}-c_{1} {\mathrm e}^{\frac {t}{2}} \sin \left (t \right )+\frac {c_{2} {\mathrm e}^{\frac {t}{2}} \sin \left (t \right )}{2}+c_{2} {\mathrm e}^{\frac {t}{2}} \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=\frac {c_{1}}{2}+c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =2\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right ) \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 4.89 (sec). Leaf size: 15
\[
y \left (t \right ) = 2 \,{\mathrm e}^{\frac {t}{2}} \left (\cos \left (t \right )+\sin \left (t \right )\right )
\]
✓ Solution by Mathematica
Time used: 0.031 (sec). Leaf size: 19
\[
y(t)\to 2 e^{t/2} (\sin (t)+\cos (t))
\]
3.2.2 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([4*diff(y(t),t$2)-4*diff(y(t),t)+5*y(t)=0,y(0) = 2, D(y)(0) = 3],y(t), singsol=all)
DSolve[{4*y''[t]-4*y'[t]+5*y[t]==0,{y[0]==2,y'[0]==3}},y[t],t,IncludeSingularSolutions -> True]