Internal problem ID [6509]
Internal file name [OUTPUT/5757_Sunday_June_05_2022_03_52_59_PM_20889182/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 7. Laplace Transforms. Section 7.5 Problesm for review and discovery. Section
A, Drill exercises. Page 309
Problem number: 4(a).
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]]
\[ \boxed {y^{\prime \prime }-5 y^{\prime }+4 y=0} \] Since no initial conditions are explicitly given, then let \begin {align*} y \left (0\right )&= c_{1} \\ y'(0) &= c_{2} \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 )-5 s Y \left (s \right )+5 y \left (0\right )+4 Y \left (s \right ) = 0\tag {1} \end {align*}
But the initial conditions are \begin {align*} y \left (0\right )&=c_{1}\\ y'(0) &=c_{2} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} s^{2} Y \left (s \right )-c_{2} -s c_{1} -5 s Y \left (s \right )+5 c_{1} +4 Y \left (s \right ) = 0 \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s c_{1} -5 c_{1} +c_{2}}{s^{2}-5 s +4} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {-\frac {c_{1}}{3}+\frac {c_{2}}{3}}{s -4}+\frac {\frac {4 c_{1}}{3}-\frac {c_{2}}{3}}{s -1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {-\frac {c_{1}}{3}+\frac {c_{2}}{3}}{s -4}\right ) &= \frac {{\mathrm e}^{4 t} \left (-c_{1} +c_{2} \right )}{3}\\ \mathcal {L}^{-1}\left (\frac {\frac {4 c_{1}}{3}-\frac {c_{2}}{3}}{s -1}\right ) &= \frac {\left (4 c_{1} -c_{2} \right ) {\mathrm e}^{t}}{3} \end {align*}
Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{\frac {5 t}{2}} \left (3 c_{1} \cosh \left (\frac {3 t}{2}\right )+\sinh \left (\frac {3 t}{2}\right ) \left (-5 c_{1} +2 c_{2} \right )\right )}{3} \] Simplifying the solution gives \[ y = \frac {\left (-5 c_{1} +2 c_{2} \right ) {\mathrm e}^{\frac {5 t}{2}} \sinh \left (\frac {3 t}{2}\right )}{3}+c_{1} {\mathrm e}^{\frac {5 t}{2}} \cosh \left (\frac {3 t}{2}\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-5 c_{1} +2 c_{2} \right ) {\mathrm e}^{\frac {5 t}{2}} \sinh \left (\frac {3 t}{2}\right )}{3}+c_{1} {\mathrm e}^{\frac {5 t}{2}} \cosh \left (\frac {3 t}{2}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-5 c_{1} +2 c_{2} \right ) {\mathrm e}^{\frac {5 t}{2}} \sinh \left (\frac {3 t}{2}\right )}{3}+c_{1} {\mathrm e}^{\frac {5 t}{2}} \cosh \left (\frac {3 t}{2}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-5 y^{\prime }+4 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-5 r +4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -1\right ) \left (r -4\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (1, 4\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )={\mathrm e}^{4 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{4 t} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 1.687 (sec). Leaf size: 33
dsolve(diff(y(t),t$2)-5*diff(y(t),t)+4*y(t)=0,y(t), singsol=all)
\[ y \left (t \right ) = \frac {{\mathrm e}^{\frac {5 t}{2}} \left (3 y \left (0\right ) \cosh \left (\frac {3 t}{2}\right )+\sinh \left (\frac {3 t}{2}\right ) \left (2 D\left (y \right )\left (0\right )-5 y \left (0\right )\right )\right )}{3} \]
✓ Solution by Mathematica
Time used: 0.012 (sec). Leaf size: 20
DSolve[y''[t]-5*y'[t]+4*y[t]==0,y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^t \left (c_2 e^{3 t}+c_1\right ) \]