2.3.6 Problem 6

Internal problem ID [9063]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 6
Date solved : Wednesday, March 05, 2025 at 07:18:53 AM
CAS classification : [_separable]

Solve

\begin{align*} t y^{\prime }+y&=0 \end{align*}

Since no initial condition is explicitly given, then let

\begin{align*} y \left (0\right ) = c_1 \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*} t^{n} f \left (t \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 \(f \left (t \right )\). Applying the above property to each term of the ode gives

\begin{align*} y &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ t y^{\prime } &\xrightarrow {\mathscr {L}} -Y \left (s \right )-s \left (\frac {d}{d s}Y \left (s \right )\right ) \end{align*}

Collecting all the terms above, the ode in Laplace domain becomes

\[ -s Y^{\prime } = 0 \]

The above ode in Y(s) is now solved.

Since the ode has the form \(Y^{\prime }=f(s)\), then we only need to integrate \(f(s)\).

\begin{align*} \int {dY} &= \int {0\, ds} + c_2 \\ Y &= c_2 \end{align*}

Applying inverse Laplace transform on the above gives.

\begin{align*} y = c_2 \delta \left (t \right )\tag {1} \end{align*}

Substituting initial conditions \(y \left (0\right ) = c_1\) and \(y^{\prime }\left (0\right ) = c_1\) into the above solution Gives

\[ c_1 = c_2 \delta \left (0\right ) \]

Solving for the constant \(c_2\) from the above equation gives

\begin{align*} c_2 = \frac {c_1}{\delta \left (0\right )} \end{align*}

Substituting the above back into the solution (1) gives

\[ y = \frac {c_1 \delta \left (t \right )}{\delta \left (0\right )} \]

Figure 2.78: Slope field \(t y^{\prime }+y = 0\)

✓ Maple. Time used: 1.020 (sec). Leaf size: 8

ode:=t*diff(y(t),t)+y(t) = 0; 
dsolve(ode,y(t),method='laplace');
 

\[ y = \delta \left (t \right ) c_{1} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & t y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {1}{t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {y^{\prime }}{y}d t =\int -\frac {1}{t}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-\ln \left (t \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{\mathit {C1}}}{t} \end {array} \]

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 16

ode=t*D[y[t],t]+y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)\to \frac {c_1}{t} \\ y(t)\to 0 \\ \end{align*}

✓ Sympy. Time used: 0.106 (sec). Leaf size: 5

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + y(t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \frac {C_{1}}{t} \]