2.3.7 Problem 7

Internal problem ID [9063]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 7
Date solved : Friday, April 25, 2025 at 05:39:52 PM
CAS classification : [_separable]

Solve

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

With initial conditions

$$\begin{array}{rl}y\left(1\right)& =5\end{array}$$

Since initial condition is not at zero, then change of variable is used to transform the ode so that initial condition is at zero.

$$\begin{array}{r}\tau =t-1\end{array}$$

Solve

$$\begin{array}{rl}(\tau +1)y^{\prime }+y& =0\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(0\right)& =5\end{array}$$

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

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

Where in the above $F(s)$ is the laplace transform of $f\left(\tau \right)$. Applying the above property to each term of the ode gives

$$\begin{array}{rl}y\left(\tau \right)& \stackrel{\mathcal{L}}{\to }Y\left(s\right)\\ (\tau +1)\left(\frac{d}{d\tau }y\left(\tau \right)\right)& \stackrel{\mathcal{L}}{\to }-Y\left(s\right)-s\left(\frac{d}{ds}Y\left(s\right)\right)+sY\left(s\right)-y\left(0\right)\end{array}$$

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

$$-s{Y}^{\prime }+sY-y\left(0\right)=0$$

Replacing $y\left(0\right)=5$ in the above results in

$$-s{Y}^{\prime }+sY-5=0$$

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

In canonical form a linear first order is

$$\begin{array}{rl}{Y}^{\prime }+q(s)Y& =p(s)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(s)& =-1\\ p(s)& =-\frac{5}{s}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qds}\\ & ={\mathrm{e}}^{\int (-1)ds}\\ & ={\mathrm{e}}^{-s}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\left(\mu \right)(-\frac{5}{s})\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(Y{\mathrm{e}}^{-s}\right)& =\left({\mathrm{e}}^{-s}\right)(-\frac{5}{s})\\ \mathrm{d}\left(Y{\mathrm{e}}^{-s}\right)& =(-\frac{5{\mathrm{e}}^{-s}}{s})\mathrm{d}s\end{array}$$

Integrating gives

$$\begin{array}{rl}Y{\mathrm{e}}^{-s}& =\int -\frac{5{\mathrm{e}}^{-s}}{s}ds\\ & =5{\mathrm{Ei}}_1\left(s\right)+c_1\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^{-s}$ gives the final solution

$$Y={\mathrm{e}}^s(5{\mathrm{Ei}}_1\left(s\right)+c_1)$$

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=\frac{5}{\tau +1}+c_1{\mathcal{L}}^{-1}({\mathrm{e}}^s,s,\tau )\end{array}$$

Substituting initial conditions $y\left(0\right)=5$ and $y^{\prime }\left(0\right)=5$ into the above solution Gives

$$5=c_1{\mathcal{L}}^{-1}({\mathrm{e}}^s,s,\tau )+5$$

Solving for the constant $c_1$ from the above equation gives

$$\begin{array}{r}{c}_1=0\end{array}$$

Substituting the above back into the solution (1) gives

$$y=\frac{5}{\tau +1}$$

Changing back the solution from $\tau$ to $t$ using

$$\begin{array}{r}\tau =t-1\end{array}$$

the solution becomes

$$\begin{array}{r}y\left(t\right)=\frac{5}{t}\end{array}$$

Solution plot	Slope field $t\left(\frac{d}{dt}y\left(t\right)\right)+y\left(t\right)=0$

✓ Maple. Time used: 0.120 (sec). Leaf size: 9

ode:=diff(y(t),t)*t+y(t) = 0; 
ic:=y(1) = 5; 
dsolve([ode,ic],y(t),method='laplace');
 

$$y=\frac{5}{t}$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [t\left(\frac{d}{dt}y\left(t\right)\right)+y\left(t\right)=0,y\left(1\right)=5]\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}=-\frac{1}{t}\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}dt=\int -\frac{1}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \ln \left(y\left(t\right)\right)=-\ln \left(t\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(t\right)\\ & & y\left(t\right)=\frac{{\mathrm{e}}^{\mathit{C}\mathit{1}}}{t}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(t\right)=\frac{\mathit{C}\mathit{1}}{t}\\ \bullet & & \text{Use initial condition}y\left(1\right)=5\\ & & 5=\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}\mathit{\text{\_C1}}\\ & & \mathit{C}\mathit{1}=5\\ \bullet & & \text{Substitute}\mathit{\text{\_C1}}=5\text{into general solution and simplify}\\ & & y\left(t\right)=\frac{5}{t}\\ \bullet & & \text{Solution to the IVP}\\ & & y\left(t\right)=\frac{5}{t}\end{array}$$

✓ Mathematica. Time used: 0.031 (sec). Leaf size: 10

ode=t*D[y[t],t]+y[t]==0; 
ic=y[1]==5; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to \frac{5}{t}$$

✓ Sympy. Time used: 0.110 (sec). Leaf size: 5

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(t*Derivative(y(t), t) + y(t),0) 
ics = {y(1): 5} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=\frac{5}{t}$$