2.3.12 Problem 12

Internal problem ID [10343]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 12
Date solved : Monday, December 08, 2025 at 08:05:57 PM
CAS classification : [_linear]

\begin{align*} \left (a t +1\right ) y^{\prime }+y&=t \\ y \left (1\right ) &= 0 \\ \end{align*}

Using Laplace transform method.

Entering first order ode laplace time varying solverSince initial condition is not at zero, then change of variable is used to transform the ode so that initial condition is at zero.

\begin{align*} \tau = t -1 \end{align*}

Entering first order ode laplace time varying solverWe 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{align*} \tau ^{n} f \left (\tau \right ) &\xrightarrow {\mathscr {L}} (-1)^n \frac {d^n}{ds^n} F(s) \end{align*}

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{align*} y &\xrightarrow {\mathscr {L}} Y \left (s \right )\\ \left (a \tau +a +1\right ) y^{\prime } &\xrightarrow {\mathscr {L}} -a \left (Y \left (s \right )+s \left (\frac {d}{d s}Y \left (s \right )\right )\right )+a \left (s Y \left (s \right )-y \left (0\right )\right )+s Y \left (s \right )-y \left (0\right )\\ \tau +1 &\xrightarrow {\mathscr {L}} \frac {1+s}{s^{2}} \end{align*}

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

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

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

\[ Y-a \left (Y+s Y^{\prime }\right )+a s Y+s Y = \frac {1+s}{s^{2}} \]

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

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} Y^{\prime } + q(s)Y &= p(s) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(s) &=-\frac {a \left (s -1\right )+s +1}{a s}\\ p(s) &=\frac {-s -1}{s^{3} a} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int -\frac {a \left (s -1\right )+s +1}{a s}d s}\\ &= s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}}\left ( \mu Y\right ) &= \left (\mu \right ) \left (\frac {-s -1}{s^{3} a}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}\right ) &= \left (s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}\right ) \left (\frac {-s -1}{s^{3} a}\right ) \\ \mathrm {d} \left (Y \,s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}\right ) &= \left (\frac {\left (-s -1\right ) s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}}{s^{3} a}\right )\, \mathrm {d} s \\ \end{align*}

Integrating gives

\begin{align*} Y \,s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}&= \int {\frac {\left (-s -1\right ) s^{\frac {a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}}{s^{3} a} \,ds} \\ &=\frac {s^{\frac {-a -1}{a}} {\mathrm e}^{-\frac {s \left (a +1\right )}{a}}}{a +1} + c_1 \end{align*}

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

\[ Y = \frac {1}{s^{2} \left (a +1\right )}+{\mathrm e}^{\frac {s \left (a +1\right )}{a}} c_1 \,s^{-\frac {a -1}{a}} \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {\tau }{a +1}+c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{s +\frac {-\left (a -1\right ) \ln \left (s \right )+s}{a}}, s , \tau \right )\tag {1} \end{align*}

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

\[ 0 = c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{s +\frac {-\left (a -1\right ) \ln \left (s \right )+s}{a}}, s , \tau \right ) \]

Solving for the constant \(c_1\) from the above equation gives

\begin{align*} c_1 = 0 \end{align*}

Substituting the above back into the solution (1) gives

\[ y = \frac {\tau }{a +1} \]

Changing back the solution from \(\tau \) to \(t\) using

\begin{align*} \tau = t -1 \end{align*}

the solution becomes

\begin{align*} y \left (t \right ) = \frac {t -1}{a +1} \end{align*}

2.3.12.1 ✓ Maple. Time used: 0.102 (sec). Leaf size: 13

ode:=(a*t+1)*diff(y(t),t)+y(t) = t; 
ic:=[y(1) = 0]; 
dsolve([ode,op(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} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (a t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right )=t , y \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d t}y \left (t \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=\frac {-y \left (t \right )+t}{a t +1} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=-\frac {y \left (t \right )}{a t +1}+\frac {t}{a t +1} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (t \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{a t +1}=\frac {t}{a t +1} \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (\frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{a t +1}\right )=\frac {\mu \left (t \right ) t}{a t +1} \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \left (t \right ) \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (\frac {d}{d t}y \left (t \right )+\frac {y \left (t \right )}{a t +1}\right )=\left (\frac {d}{d t}y \left (t \right )\right ) \mu \left (t \right )+y \left (t \right ) \left (\frac {d}{d t}\mu \left (t \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d t}\mu \left (t \right ) \\ {} & {} & \frac {d}{d t}\mu \left (t \right )=\frac {\mu \left (t \right )}{a t +1} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )=\left (a t +1\right )^{\frac {1}{a}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \left (t \right ) \mu \left (t \right )\right )\right )d t =\int \frac {\mu \left (t \right ) t}{a t +1}d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \left (t \right ) \mu \left (t \right )=\int \frac {\mu \left (t \right ) t}{a t +1}d t +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\frac {\int \frac {\mu \left (t \right ) t}{a t +1}d t +\mathit {C1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )=\left (a t +1\right )^{\frac {1}{a}} \\ {} & {} & y \left (t \right )=\frac {\int \frac {\left (a t +1\right )^{\frac {1}{a}} t}{a t +1}d t +\mathit {C1}}{\left (a t +1\right )^{\frac {1}{a}}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\frac {\left (t -1\right ) \left (a t +1\right )^{\frac {1}{a}}}{a +1}+\mathit {C1}}{\left (a t +1\right )^{\frac {1}{a}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t -1+\left (a t +1\right )^{-\frac {1}{a}} \mathit {C1} \left (a +1\right )}{a +1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=\left (a +1\right )^{-\frac {1}{a}} \mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t -1}{a +1} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t -1}{a +1} \end {array} \]

2.3.12.2 ✓ Mathematica. Time used: 0.569 (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]
 

\begin{align*} y(t)&\to \frac {t-1}{a+1} \end{align*}

2.3.12.3 ✗ 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)