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 : Sunday, March 30, 2025 at 02:06:00 PM
CAS classification : [_separable]

Solve

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

Since no initial condition is explicitly given, then let

$$\begin{array}{r}y\left(0\right)=c_1\end{array}$$

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 $f(s)$.

$$\begin{array}{rl}\int dY& =\int 0ds+c_2\\ Y& =c_2\end{array}$$

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=c_2\delta \left(t\right)\end{array}$$

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

$$\begin{array}{r}{c}_2=\frac{c_1}{\delta \left(0\right)}\end{array}$$

Substituting the above back into the solution (1) gives

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

Figure 2.73: Slope field $t{y}^{\prime }+y=0$

✓ Maple. Time used: 0.077 (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

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

✓ 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]
 

$$\begin{array}{r}y(t)\to \frac{c_1}{t}\\ y(t)\to 0\end{array}$$

✓ 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)
 

$$y\left(t\right)=\frac{C_1}{t}$$