2.3.12 Problem 12

Internal problem ID [9068]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 12
Date solved : Friday, April 25, 2025 at 05:40:00 PM
CAS classification : [_linear]

Solve

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

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

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

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

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

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

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

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

$$-a(Y+s{Y}^{\prime })+asY+sY+Y=\frac{1+s}{s^2}$$

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

$$\begin{array}{rl}q(s)& =-\frac{(s-1)a+1+s}{as}\\ p(s)& =\frac{-s-1}{s^{3}a}\end{array}$$

The integrating factor $\mu$ is

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

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(\mu Y\right)& =\left(\mu \right)\left(\frac{-s-1}{s^{3}a}\right)\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(Y{s}^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}\right)& =\left(s^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}\right)\left(\frac{-s-1}{s^{3}a}\right)\\ \mathrm{d}\left(Y{s}^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}\right)& =\left(\frac{(-s-1)s^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}}{s^{3}a}\right)\mathrm{d}s\end{array}$$

Integrating gives

$$\begin{array}{rl}Y{s}^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}& =\int \frac{(-s-1)s^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}}{s^{3}a}ds\\ & =\frac{s^{\frac{-a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}}{a+1}+c_1\end{array}$$

Dividing throughout by the integrating factor $s^{\frac{a-1}{a}}{\mathrm{e}}^{-\frac{s(a+1)}{a}}$ gives the final solution

$$Y=\frac{1}{s^2(a+1)}+{\mathrm{e}}^{\frac{s(a+1)}{a}}{c}_1{s}^{-\frac{a-1}{a}}$$

Applying inverse Laplace transform on the above gives.

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

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

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

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

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

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

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

the solution becomes

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

✓ Maple. Time used: 0.139 (sec). Leaf size: 13

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

$$y=\frac{t-1}{a+1}$$

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}\\ & & [(at+1)\left(\frac{d}{dt}y\left(t\right)\right)+y\left(t\right)=t,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{Solve for the highest derivative}\\ & & \frac{d}{dt}y\left(t\right)=\frac{-y\left(t\right)+t}{at+1}\\ \bullet & & \text{Collect w.r.t.}y\left(t\right)\text{and simplify}\\ & & \frac{d}{dt}y\left(t\right)=-\frac{y\left(t\right)}{at+1}+\frac{t}{at+1}\\ \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)}{at+1}=\frac{t}{at+1}\\ \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)}{at+1})=\frac{\mu \left(t\right)t}{at+1}\\ \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)}{at+1})=\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)}{at+1}\\ \bullet & & \text{Solve to find the integrating factor}\\ & & \mu \left(t\right)={(at+1)}^{\frac{1}{a}}\\ \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)t}{at+1}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)t}{at+1}dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(t\right)\\ & & y\left(t\right)=\frac{\int \frac{\mu \left(t\right)t}{at+1}dt+\mathit{C}\mathit{1}}{\mu \left(t\right)}\\ \bullet & & \text{Substitute}\mu \left(t\right)={(at+1)}^{\frac{1}{a}}\\ & & y\left(t\right)=\frac{\int \frac{{(at+1)}^{\frac{1}{a}}t}{at+1}dt+\mathit{C}\mathit{1}}{{(at+1)}^{\frac{1}{a}}}\\ \bullet & & \text{Evaluate the integrals on the rhs}\\ & & y\left(t\right)=\frac{\frac{(t-1){(at+1)}^{\frac{1}{a}}}{a+1}+\mathit{C}\mathit{1}}{{(at+1)}^{\frac{1}{a}}}\\ \bullet & & \text{Simplify}\\ & & y\left(t\right)=\frac{t-1+{(at+1)}^{-\frac{1}{a}}\mathit{C}\mathit{1}(a+1)}{a+1}\\ \bullet & & \text{Use initial condition}y\left(1\right)=0\\ & & 0={(a+1)}^{-\frac{1}{a}}\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}\mathit{\text{\_C1}}\\ & & \mathit{C}\mathit{1}=0\\ \bullet & & \text{Substitute}\mathit{\text{\_C1}}=0\text{into general solution and simplify}\\ & & y\left(t\right)=\frac{t-1}{a+1}\\ \bullet & & \text{Solution to the IVP}\\ & & y\left(t\right)=\frac{t-1}{a+1}\end{array}$$

✓ Mathematica. Time used: 0.599 (sec). Leaf size: 14

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

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

✗ Sympy

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