2.3.4 Problem 4

Internal problem ID [10335]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 4
Date solved : Thursday, November 27, 2025 at 10:34:33 AM
CAS classification : [_separable]

\begin{align*} y^{\prime } t +y&=0 \\ y \left (0\right ) &= y_{0} \\ \end{align*}

Using Laplace transform method.

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.

Solve 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_1 \\ Y &= c_1 \end{align*}

Applying inverse Laplace transform on the above gives.

\begin{align*} y = c_1 \delta \left (t \right )\tag {1} \end{align*}

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

\[ y_{0} = c_1 \delta \left (0\right ) \]

Solving for the constant \(c_1\) from the above equation gives

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

Substituting the above back into the solution (1) gives

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

Figure 2.81: Slope field \(y^{\prime } t +y = 0\)

✓ Maple. Time used: 0.059 (sec). Leaf size: 12

ode:=t*diff(y(t),t)+y(t) = 0; 
ic:=[y(0) = y__0]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

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

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

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

✗ Mathematica

ode=t*D[y[t],t]+y[t]==0; 
ic=y[0]==y0; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

Not solved

✓ Sympy. Time used: 0.065 (sec). Leaf size: 3

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + y(t),0) 
ics = {y(0): y__0} 
dsolve(ode,func=y(t),ics=ics)
 

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