2.3.6 Problem 6

Internal problem ID [10337]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 6
Date solved : Monday, January 26, 2026 at 09:43:35 PM
CAS classification : [_separable]

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

Using Laplace transform method.

Entering first order ode laplace time varying solverSince 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 )\\ y^{\prime } t &\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.

Entering first order ode quadrature solverSince 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

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

Direction field \(y^{\prime } t +y = 0\)	Isoclines for \(y^{\prime } t +y = 0\)

2.3.6.1 ✓ Maple. Time used: 0.056 (sec). Leaf size: 8

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

\[ y = c_1 \delta \left (t \right ) \]

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

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

2.3.6.2 ✓ Mathematica. Time used: 0.015 (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*}

2.3.6.3 ✓ Sympy. Time used: 0.122 (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)
 

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

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(t)) 
 
('factorable', 'separable', '1st_exact', '1st_linear', 'Bernoulli', '1st_homogeneous_coeff_best', '1st_homogeneous_coeff_subs_indep_div_dep', '1st_homogeneous_coeff_subs_dep_div_indep', 'almost_linear', 'lie_group', 'nth_linear_euler_eq_homogeneous', 'separable_Integral', '1st_exact_Integral', '1st_linear_Integral', 'Bernoulli_Integral', '1st_homogeneous_coeff_subs_indep_div_dep_Integral', '1st_homogeneous_coeff_subs_dep_div_indep_Integral', 'almost_linear_Integral')