2.3.14 Problem 14

Internal problem ID [9070]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 14
Date solved : Friday, April 25, 2025 at 05:40:03 PM
CAS classification : [_separable]

Solve

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

With initial conditions

$$\begin{array}{rl}y(-3)& =0\end{array}$$

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

$$\begin{array}{r}\tau =t+3\end{array}$$

Solve

$$\begin{array}{rl}{y}^{\prime }+(a(\tau -3)+b(\tau -3))y& =0\end{array}$$

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

$$\begin{array}{rl}(a\tau +b\tau -3a-3b)y\left(\tau \right)& \stackrel{\mathcal{L}}{\to }-a\left(\frac{d}{ds}Y\left(s\right)\right)-b\left(\frac{d}{ds}Y\left(s\right)\right)-3aY\left(s\right)-3bY\left(s\right)\\ \frac{d}{d\tau }y\left(\tau \right)& \stackrel{\mathcal{L}}{\to }Y\left(s\right)s-y\left(0\right)\end{array}$$

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

$$Ys-y\left(0\right)-a{Y}^{\prime }-b{Y}^{\prime }-3aY-3bY=0$$

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

$$Ys-a{Y}^{\prime }-b{Y}^{\prime }-3aY-3bY=0$$

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

$$\begin{array}{rl}q(s)& =-\frac{-3a-3b+s}{a+b}\\ p(s)& =0\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qds}\\ & ={\mathrm{e}}^{\int -\frac{-3a-3b+s}{a+b}ds}\\ & ={\mathrm{e}}^{\frac{s(6a+6b-s)}{2a+2b}}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}s}\mu Y& =0\\ \frac{\mathrm{d}}{\mathrm{d}s}\left(Y{\mathrm{e}}^{\frac{s(6a+6b-s)}{2a+2b}}\right)& =0\end{array}$$

Integrating gives

$$\begin{array}{rl}Y{\mathrm{e}}^{\frac{s(6a+6b-s)}{2a+2b}}& =\int 0ds+c_1\\ & =c_1\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^{\frac{s(6a+6b-s)}{2a+2b}}$ gives the final solution

$$Y=c_1{\mathrm{e}}^{-\frac{s(6a+6b-s)}{2a+2b}}$$

Applying inverse Laplace transform on the above gives.

$$\begin{array}{}\text{(1)}& y=c_1{\mathcal{L}}^{-1}({\mathrm{e}}^{-\frac{s(6a+6b-s)}{2a+2b}},s,\tau )\end{array}$$

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

$$0=c_1{\mathcal{L}}^{-1}({\mathrm{e}}^{-\frac{s(6a+6b-s)}{2a+2b}},s,\tau )$$

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

$$y=0$$

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

$$\begin{array}{r}\tau =t+3\end{array}$$

the solution becomes

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

Figure 2.79: Solution plot

✓ Maple. Time used: 0.107 (sec). Leaf size: 5

ode:=diff(y(t),t)+(a*t+b*t)*y(t) = 0; 
ic:=y(-3) = 0; 
dsolve([ode,ic],y(t),method='laplace');
 

$$y=0$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d}{dt}y\left(t\right)+(at+bt)y\left(t\right)=0,y(-3)=0]\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dt}y\left(t\right)=-(at+bt)y\left(t\right)\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}=-at-bt\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \frac{\frac{d}{dt}y\left(t\right)}{y\left(t\right)}dt=\int (-at-bt)dt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & \ln \left(y\left(t\right)\right)=-\frac{t^2(a+b)}{2}+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}y\left(t\right)\\ & & y\left(t\right)={\mathrm{e}}^{-\frac{1}{2}{t}^{2}a-\frac{1}{2}{t}^{2}b+\mathit{C}\mathit{1}}\\ \bullet & & \text{Simplify}\\ & & y\left(t\right)={\mathrm{e}}^{\frac{(-a-b)t^2}{2}+\mathit{C}\mathit{1}}\\ \bullet & & \text{Redefine the integration constant(s)}\\ & & y\left(t\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{\frac{(-a-b)t^2}{2}}\\ \bullet & & \text{Use initial condition}y(-3)=0\\ & & 0=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{9a}{2}-\frac{9b}{2}}\\ \bullet & & \text{Solve for}\mathit{\text{\_C1}}\\ & & \mathit{C}\mathit{1}=0\\ \bullet & & \text{Substitute}\mathit{\text{\_C1}}=0\text{into general solution and simplify}\\ & & y\left(t\right)=0\\ \bullet & & \text{Solution to the IVP}\\ & & y\left(t\right)=0\end{array}$$

✓ Mathematica. Time used: 0.002 (sec). Leaf size: 6

ode=D[y[t],t]+(a*t+b*t)*y[t]==0; 
ic=y[-3]==0; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to 0$$

✓ Sympy. Time used: 0.326 (sec). Leaf size: 3

from sympy import * 
t = symbols("t") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq((a*t + b*t)*y(t) + Derivative(y(t), t),0) 
ics = {y(-3): 0} 
dsolve(ode,func=y(t),ics=ics)
 

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