2.3.10 Problem 10

Internal problem ID [10341]
Book : First order enumerated odes
Section : section 3. First order odes solved using Laplace method
Problem number : 10
Date solved : Thursday, November 27, 2025 at 10:34:43 AM
CAS classification : [_linear]

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

Using Laplace transform method.

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

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

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

Replacing \(y \left (0\right ) = 1\) in the above results in

\[ -s Y^{\prime }+s Y-1 = \frac {1+s}{s^{2}} \]

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) &=-1\\ p(s) &=\frac {-s^{2}-s -1}{s^{3}} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,ds}}\\ &= {\mathrm e}^{\int \left (-1\right )d s}\\ &= {\mathrm e}^{-s} \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^{2}-s -1}{s^{3}}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}s}} \left (Y \,{\mathrm e}^{-s}\right ) &= \left ({\mathrm e}^{-s}\right ) \left (\frac {-s^{2}-s -1}{s^{3}}\right ) \\ \mathrm {d} \left (Y \,{\mathrm e}^{-s}\right ) &= \left (\frac {\left (-s^{2}-s -1\right ) {\mathrm e}^{-s}}{s^{3}}\right )\, \mathrm {d} s \\ \end{align*}

Integrating gives

\begin{align*} Y \,{\mathrm e}^{-s}&= \int {\frac {\left (-s^{2}-s -1\right ) {\mathrm e}^{-s}}{s^{3}} \,ds} \\ &=\frac {{\mathrm e}^{-s}}{2 s}+\frac {\operatorname {Ei}_{1}\left (s \right )}{2}+\frac {{\mathrm e}^{-s}}{2 s^{2}} + c_1 \end{align*}

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

\[ Y = {\mathrm e}^{s} \left (\frac {\operatorname {Ei}_{1}\left (s \right )}{2}+c_1 \right )+\frac {1}{2 s}+\frac {1}{2 s^{2}} \]

Applying inverse Laplace transform on the above gives.

\begin{align*} y = \frac {1}{2 \tau +2}+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}, s , \tau \right )+\frac {1}{2}+\frac {\tau }{2}\tag {1} \end{align*}

Substituting initial conditions \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = 1\) into the above solution Gives

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

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

\[ y = \frac {1}{2 \tau +2}+\frac {1}{2}+\frac {\tau }{2} \]

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 {\left (t -1\right )^{2}+2 t}{2 t} \end{align*}

Which simplifies to

\[ y \left (t \right ) = \frac {t^{2}+1}{2 t} \]

Solution plot	Slope field \(\left (\frac {d}{d t}y \left (t \right )\right ) t +y \left (t \right ) = t\)

✓ Maple. Time used: 0.091 (sec). Leaf size: 13

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

\[ y = \frac {t}{2}+\frac {1}{2 t} \]

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 [t \left (\frac {d}{d t}y \left (t \right )\right )+y \left (t \right )=t , y \left (1\right )=1\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 {Isolate the derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y \left (t \right )=1-\frac {y \left (t \right )}{t} \\ \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 )}{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 )}{t}\right )=\mu \left (t \right ) \\ \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 )}{t}\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 )}{t} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )=t \\ \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 \mu \left (t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \left (t \right ) \mu \left (t \right )=\int \mu \left (t \right )d t +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\frac {\int \mu \left (t \right )d t +\mathit {C1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )=t \\ {} & {} & y \left (t \right )=\frac {\int t d t +\mathit {C1}}{t} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\frac {t^{2}}{2}+\mathit {C1}}{t} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t^{2}+2 \mathit {C1}}{2 t} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\mathit {C1} +\frac {1}{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_C1} \\ {} & {} & \mathit {C1} =\frac {1}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_C1} =\frac {1}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t^{2}+1}{2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {t^{2}+1}{2 t} \end {array} \]

✓ Mathematica. Time used: 0.017 (sec). Leaf size: 17

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

\begin{align*} y(t)&\to \frac {t^2+1}{2 t} \end{align*}

✓ Sympy. Time used: 0.102 (sec). Leaf size: 10

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

\[ y{\left (t \right )} = \frac {t}{2} + \frac {1}{2 t} \]