Internal problem ID [5680]
Internal file name [OUTPUT/4928_Sunday_June_05_2022_03_10_51_PM_30685895/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number: 2.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {2 y+y^{\prime }=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {3}{2}}\right ] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}
Where here \begin {align*} p(t) &=2\\ q(t) &=0 \end {align*}
Hence the ode is \begin {align*} 2 y+y^{\prime } = 0 \end {align*}
The domain of \(p(t)=2\) is \[
\{-\infty
Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] 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 ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} 2 Y \left (s \right )+s Y \left (s \right )-y \left (0\right ) = 0\tag {1} \end {align*}
Replacing initial condition gives \begin {align*} 2 Y \left (s \right )+s Y \left (s \right )-\frac {3}{2} = 0 \end {align*}
Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {3}{2 \left (2+s \right )} \end {align*}
Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {3}{2 \left (2+s \right )}\right )\\ &= \frac {3 \,{\mathrm e}^{-2 t}}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {3 \,{\mathrm e}^{-2 t}}{2} \\
\end{align*} Verification of solutions
\[
y = \frac {3 \,{\mathrm e}^{-2 t}}{2}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y+y^{\prime }=0, y \left (0\right )=\frac {3}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-2 y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-2 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int \left (-2\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-2 t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{-2 t +c_{1}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {3}{2} \\ {} & {} & \frac {3}{2}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\ln \left (\frac {3}{2}\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\ln \left (\frac {3}{2}\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-2 t}}{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {3 \,{\mathrm e}^{-2 t}}{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.688 (sec). Leaf size: 10
\[
y \left (t \right ) = \frac {3 \,{\mathrm e}^{-2 t}}{2}
\]
✓ Solution by Mathematica
Time used: 0.038 (sec). Leaf size: 31
\[
y(t)\to \frac {1}{4} \left (11 e^{-26 t/5}+13 \sin (2 t)-5 \cos (2 t)\right )
\]
6.2.2 Solving as laplace ode
6.2.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
dsolve([diff(y(t),t)+2*y(t)=0,y(0) = 3/2],y(t), singsol=all)
DSolve[{y'[t]+52/10*y[t]==194/10*Sin[2*t],{y[0]==15/10}},y[t],t,IncludeSingularSolutions -> True]