2.3.8 Problem 8

Internal problem ID [9064]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 8
Date solved : Friday, April 25, 2025 at 05:39:54 PM
CAS classification : [_linear]

Solve

$$\begin{array}{rl}t{y}^{\prime }+y& =\sin \left(t\right)\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(1\right)& =0\end{array}$$

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

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

With initial conditions

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

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

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

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

$$\begin{array}{rl}y\left(\tau \right)& \stackrel{\mathcal{L}}{\to }Y\left(s\right)\\ (\tau +1)\left(\frac{d}{d\tau }y\left(\tau \right)\right)& \stackrel{\mathcal{L}}{\to }-Y\left(s\right)-s\left(\frac{d}{ds}Y\left(s\right)\right)+sY\left(s\right)-y\left(0\right)\\ \sin (\tau +1)& \stackrel{\mathcal{L}}{\to }\frac{\sin \left(1\right)s+\cos \left(1\right)}{s^2+1}\end{array}$$

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

$$-s{Y}^{\prime }+sY-y\left(0\right)=\frac{\sin \left(1\right)s+\cos \left(1\right)}{s^2+1}$$

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

$$-s{Y}^{\prime }+sY=\frac{\sin \left(1\right)s+\cos \left(1\right)}{s^2+1}$$

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

$$\begin{array}{rl}q(s)& =-1\\ p(s)& =\frac{-\sin \left(1\right)s-\cos \left(1\right)}{(s^2+1)s}\end{array}$$

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

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\left(\mu \right)\left(\frac{-\sin \left(1\right)s-\cos \left(1\right)}{(s^2+1)s}\right)\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(Y{\mathrm{e}}^{-s}\right)& =\left({\mathrm{e}}^{-s}\right)\left(\frac{-\sin \left(1\right)s-\cos \left(1\right)}{(s^2+1)s}\right)\\ \mathrm{d}\left(Y{\mathrm{e}}^{-s}\right)& =\left(\frac{(-\sin \left(1\right)s-\cos \left(1\right)){\mathrm{e}}^{-s}}{(s^2+1)s}\right)\mathrm{d}s\end{array}$$

Integrating gives

$$\begin{array}{rl}Y{\mathrm{e}}^{-s}& =\int \frac{(-\sin \left(1\right)s-\cos \left(1\right)){\mathrm{e}}^{-s}}{(s^2+1)s}ds\\ & =-\frac{{\mathrm{Ei}}_1(s+i)}{2}-\frac{{\mathrm{Ei}}_1(s-i)}{2}+\cos \left(1\right){\mathrm{Ei}}_1\left(s\right)+c_1\end{array}$$

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

$$Y={\mathrm{e}}^s(-\frac{{\mathrm{Ei}}_1(s+i)}{2}-\frac{{\mathrm{Ei}}_1(s-i)}{2}+\cos \left(1\right){\mathrm{Ei}}_1\left(s\right)+c_1)$$

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=\frac{\cos \left(1\right)}{\tau +1}+c_1{\mathcal{L}}^{-1}({\mathrm{e}}^s,s,\tau )-\frac{\cos (\tau +1)}{\tau +2}\end{array}$$

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

$$0=c_1{\mathcal{L}}^{-1}({\mathrm{e}}^s,s,\tau )+\frac{\cos \left(1\right)}{2}$$

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

$$\begin{array}{r}{c}_1=-\frac{\cos \left(1\right)}{2{\mathcal{L}}^{-1}({\mathrm{e}}^s,s,\tau )}\end{array}$$

Substituting the above back into the solution (1) gives

$$y=\frac{\cos \left(1\right)}{\tau +1}-\frac{\cos \left(1\right)}{2}-\frac{\cos (\tau +1)}{\tau +2}$$

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

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

the solution becomes

$$\begin{array}{r}y\left(t\right)=\frac{\cos \left(1\right)}{t}-\frac{\cos \left(1\right)}{2}-\frac{\cos \left(t\right)}{t+1}\end{array}$$

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

✗ Maple

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

$$\text{No solution found}$$

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)=\sin \left(t\right),y\left(1\right)=0]\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Isolate the derivative}\\ & & \frac{d}{dt}y\left(t\right)=-\frac{y\left(t\right)}{t}+\frac{\sin \left(t\right)}{t}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ & & \frac{d}{dt}y\left(t\right)+\frac{y\left(t\right)}{t}=\frac{\sin \left(t\right)}{t}\\ \bullet & & \text{The ODE is linear; multiply by an integrating factor}\mu \left(t\right)\\ & & \mu \left(t\right)(\frac{d}{dt}y\left(t\right)+\frac{y\left(t\right)}{t})=\frac{\mu \left(t\right)\sin \left(t\right)}{t}\\ \bullet & & \text{Assume the lhs of the ODE is the total derivative}\frac{d}{dt}\left(y\left(t\right)\mu \left(t\right)\right)\\ & & \mu \left(t\right)(\frac{d}{dt}y\left(t\right)+\frac{y\left(t\right)}{t})=\left(\frac{d}{dt}y\left(t\right)\right)\mu \left(t\right)+y\left(t\right)\left(\frac{d}{dt}\mu \left(t\right)\right)\\ \bullet & & \text{Isolate}\frac{d}{dt}\mu \left(t\right)\\ & & \frac{d}{dt}\mu \left(t\right)=\frac{\mu \left(t\right)}{t}\\ \bullet & & \text{Solve to find the integrating factor}\\ & & \mu \left(t\right)=t\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \left(\frac{d}{dt}\left(y\left(t\right)\mu \left(t\right)\right)\right)dt=\int \frac{\mu \left(t\right)\sin \left(t\right)}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate the integral on the lhs}\\ & & y\left(t\right)\mu \left(t\right)=\int \frac{\mu \left(t\right)\sin \left(t\right)}{t}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(t\right)\\ & & y\left(t\right)=\frac{\int \frac{\mu \left(t\right)\sin \left(t\right)}{t}dt+\mathit{C}\mathit{1}}{\mu \left(t\right)}\\ \bullet & & \text{Substitute}\mu \left(t\right)=t\\ & & y\left(t\right)=\frac{\int \sin \left(t\right)dt+\mathit{C}\mathit{1}}{t}\\ \bullet & & \text{Evaluate the integrals on the rhs}\\ & & y\left(t\right)=\frac{-\cos \left(t\right)+\mathit{C}\mathit{1}}{t}\\ \bullet & & \text{Use initial condition}y\left(1\right)=0\\ & & 0=-\cos \left(1\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}\mathit{\text{\_C1}}\\ & & \mathit{C}\mathit{1}=\cos \left(1\right)\\ \bullet & & \text{Substitute}\mathit{\text{\_C1}}=\cos \left(1\right)\text{into general solution and simplify}\\ & & y\left(t\right)=\frac{-\cos \left(t\right)+\cos \left(1\right)}{t}\\ \bullet & & \text{Solution to the IVP}\\ & & y\left(t\right)=\frac{-\cos \left(t\right)+\cos \left(1\right)}{t}\end{array}$$

✓ Mathematica. Time used: 0.035 (sec). Leaf size: 16

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

$$y(t)\to \frac{\cos (1)-\cos (t)}{t}$$

✓ Sympy. Time used: 0.252 (sec). Leaf size: 10

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

$$y\left(t\right)=\frac{-\cos \left(t\right)+\cos \left(1\right)}{t}$$