2.3.4 Problem 4

Internal problem ID [9061]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 4
Date solved : Sunday, March 30, 2025 at 02:05:57 PM
CAS classification : [_separable]

Solve

$$\begin{array}{rl}t{y}^{\prime }+y& =0\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(0\right)& =y_0\end{array}$$

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

$$\begin{array}{rl}{t}^{n}f\left(t\right)& \stackrel{\mathcal{L}}{\to }(-1{)}^n\frac{d^n}{d{s}^n}F(s)\end{array}$$

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

$$\begin{array}{rl}y& \stackrel{\mathcal{L}}{\to }Y\left(s\right)\\ t{y}^{\prime }& \stackrel{\mathcal{L}}{\to }-Y\left(s\right)-s\left(\frac{d}{ds}Y\left(s\right)\right)\end{array}$$

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 $f(s)$.

$$\begin{array}{rl}\int dY& =\int 0ds+c_1\\ Y& =c_1\end{array}$$

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=c_1\delta \left(t\right)\end{array}$$

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

$$\begin{array}{r}{c}_1=\frac{y_0}{\delta \left(0\right)}\end{array}$$

Substituting the above back into the solution (1) gives

$$y=\frac{y_0\delta \left(t\right)}{\delta \left(0\right)}$$

The solution was found not to satisfy the ode or the IC. Hence it is removed.

✓ Maple. Time used: 0.077 (sec). Leaf size: 12

ode:=diff(y(t),t)*t+y(t) = 0; 
ic:=y(0) = y__0; 
dsolve([ode,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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [t\left(\frac{d}{dt}y\left(t\right)\right)+y\left(t\right)=0,y\left(0\right)=y_0]\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}=-\frac{1}{t}\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}dt=\int -\frac{1}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \ln \left(y\left(t\right)\right)=-\ln \left(t\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(t\right)\\ & & y\left(t\right)=\frac{{\mathrm{e}}^{\mathit{C}\mathit{1}}}{t}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(t\right)=\frac{\mathit{C}\mathit{1}}{t}\\ \bullet & & \text{Solution does not satisfy initial condition}\end{array}$$

✗ 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: 0.104 (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)
 

$$y\left(t\right)=0$$