2.3.13 Problem 13

Internal problem ID [10344]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 13
Date solved : Thursday, November 27, 2025 at 10:34:47 AM
CAS classification : [_separable]

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

Using Laplace transform method.

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{align*} t^{n} f \left (t \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 \(f \left (t \right )\). Applying the above property to each term of the ode gives

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

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

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

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

\[ -a Y^{\prime }-b Y^{\prime }+Y s = 0 \]

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

Solve In 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 {s}{a +b}\\ p(s) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int -\frac {s}{a +b}d s}\\ &= {\mathrm e}^{-\frac {s^{2}}{2 a +2 b}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \mu Y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{-\frac {s^{2}}{2 a +2 b}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{-\frac {s^{2}}{2 a +2 b}}&= \int {0 \,ds} + c_1 \\ &=c_1 \end{align*}

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

\[ Y = c_1 \,{\mathrm e}^{\frac {s^{2}}{2 a +2 b}} \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = 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}^{\frac {s^{2}}{2 a +2 b}}, s , t\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

\[ 0 = c_1 \operatorname {Typesetting}\mcoloneq \operatorname {msup}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mi}\left (\text {``$\mathcal \{L\}$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mrow}\left (\operatorname {Typesetting}\mcoloneq \operatorname {mo}\left (\text {``$-$''}\right ), \operatorname {Typesetting}\mcoloneq \operatorname {mn}\left (``1''\right )\right ), \operatorname {Typesetting}\mcoloneq \operatorname {msemantics}=\text {``atomic''}\right )\left ({\mathrm e}^{\frac {s^{2}}{2 a +2 b}}, s , t\right ) \]

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

\[ y = 0 \]

Figure 2.84: Solutions plot

✓ Maple. Time used: 0.069 (sec). Leaf size: 5

ode:=diff(y(t),t)+(a*t+b*t)*y(t) = 0; 
ic:=[y(0) = 0]; 
dsolve([ode,op(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} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d t}y \left (t \right )+\left (a t +b t \right ) y \left (t \right )=0, y \left (0\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 )=-\left (a t +b t \right ) y \left (t \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d t}y \left (t \right )}{y \left (t \right )}=-a t -b t \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {\frac {d}{d t}y \left (t \right )}{y \left (t \right )}d t =\int \left (-a t -b t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y \left (t \right )\right )=-\frac {t^{2} \left (a +b \right )}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )={\mathrm e}^{-\frac {1}{2} t^{2} a -\frac {1}{2} t^{2} b +\mathit {C1}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )={\mathrm e}^{\frac {\left (-a -b \right ) t^{2}}{2}+\mathit {C1}} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{\frac {\left (-a -b \right ) t^{2}}{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=\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 )=0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 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[0]==0; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)&\to 0 \end{align*}

✓ Sympy. Time used: 0.206 (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(0): 0} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = 0 \]