2.3.7 Problem 7
Internal
problem
ID
[9064]
Book
:
First
order
enumerated
odes
Section
:
section
3.
First
order
odes
solved
using
Laplace
method
Problem
number
:
7
Date
solved
:
Friday, February 21, 2025 at 09:07:31 PM
CAS
classification
:
[_separable]
Solve
\begin{align*} t y^{\prime }+y&=0 \end{align*}
With initial conditions
\begin{align*} y \left (1\right )&=5 \end{align*}
Since initial condition is not at zero, then change of variable is used to transform the ode so that initial condition is at zero.
\begin{align*} \tau = t -1 \end{align*}
Solve
\begin{align*} \left (\tau +1\right ) y^{\prime }+y&=0 \end{align*}
With initial conditions
\begin{align*} y \left (0\right )&=5 \end{align*}
We will now apply Laplace transform to each term in the ode. Since this is time varying, the following Laplace transform property will be used
\begin{align*} \tau ^{n} f \left (\tau \right ) &\xrightarrow {\mathscr {L}} (-1)^n \frac {d^n}{ds^n} F(s) \end{align*}
Where in the above \(F(s)\) is the laplace transform of \(f \left (\tau \right )\). Applying the above property to each term of the ode gives
\begin{align*} y \left (\tau \right ) &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ \left (\tau +1\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right ) &\xrightarrow {\mathscr {L}} -Y \left (s \right )-s \left (\frac {d}{d s}Y \left (s \right )\right )+s Y \left (s \right )-y \left (0\right ) \end{align*}
Collecting all the terms above, the ode in Laplace domain becomes
\[
-s Y^{\prime }+s Y-y \left (0\right ) = 0
\]
Replacing \(y \left (0\right ) = 5\) in the above results in
\[
-s Y^{\prime }+s Y-5 = 0
\]
The above ode in Y(s) is now solved.
In canonical form a linear first order is
\begin{align*} Y^{\prime } + q(s)Y &= p(s) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(s) &=-1\\ p(s) &=-\frac {5}{s} \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int \left (-1\right )d s}\\ &= {\mathrm e}^{-s} \end{align*}
The ode becomes
\begin{align*}
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \mu p \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \left (\mu \right ) \left (-\frac {5}{s}\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{-s}\right ) &= \left ({\mathrm e}^{-s}\right ) \left (-\frac {5}{s}\right ) \\
\mathrm {d} \left (Y \,{\mathrm e}^{-s}\right ) &= \left (-\frac {5 \,{\mathrm e}^{-s}}{s}\right )\, \mathrm {d} s \\
\end{align*}
Integrating gives
\begin{align*} Y \,{\mathrm e}^{-s}&= \int {-\frac {5 \,{\mathrm e}^{-s}}{s} \,ds} \\ &=5 \,\operatorname {Ei}_{1}\left (s \right ) + c_1 \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{-s}\) gives the final solution
\[ Y = {\mathrm e}^{s} \left (5 \,\operatorname {Ei}_{1}\left (s \right )+c_1 \right ) \]
Applying inverse Laplace transform on the above gives.
\begin{align*} y = \frac {5}{\tau +1}+c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{s}, s , \tau \right )\tag {1} \end{align*}
Substituting initial conditions \(y \left (0\right ) = 5\) and \(y^{\prime }\left (0\right ) = 5\) into the above solution Gives
\[
5 = c_1 \mathcal {L}^{-1}\left ({\mathrm e}^{s}, s , \tau \right )+5
\]
Solving for the constant \(c_1\) from the above equation gives
\begin{align*} c_1 = 0 \end{align*}
Substituting the above back into the solution (1) gives
\[
y = \frac {5}{\tau +1}
\]
Changing back the solution from \(\tau \) to \(t\) using
\begin{align*} \tau = t -1 \end{align*}
the solution becomes
\begin{align*} y \left (t \right ) = \frac {5}{t} \end{align*}
|
|
|
| Solution \(y \left (t \right ) = \frac {5}{t}\) | Slope field \(t \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right ) = 0\) |
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [t y^{\prime }+y=0, y \left (1\right )=5\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {1}{t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int -\frac {1}{t}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\ln \left (t \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{\mathit {C1}}}{t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=5 \\ {} & {} & 5={\mathrm e}^{\mathit {C1}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\ln \left (5\right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\ln \left (5\right )\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {5}{t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {5}{t} \end {array} \]
Maple trace
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
Maple dsolve solution
Solving time : 1.458 (sec)
Leaf size : 9
dsolve([t*diff(y(t),t)+y(t) = 0,op([y(1) = 5])],y(t),method=laplace)
\[
y = \frac {5}{t}
\]
✓Mathematica DSolve solution
Solving time : 0.031 (sec)
Leaf size : 10
DSolve[{t*D[y[t],t]+y[t]==0,y[1]==5},y[t],t,IncludeSingularSolutions->True]
\[
y(t)\to \frac {5}{t}
\]
✓Sympy solution
Solving time : 0.110 (sec)
Leaf size : 5
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(t*Derivative(y(t), t) + y(t),0)
ics = {y(1): 5}
dsolve(ode,func=y(t),ics=ics)
Eq(y(t), 5/t)
\[
y{\left (t \right )} = \frac {5}{t}
\]